Chemistry Steps

Chemistry Steps

stoichiometry worksheet 3 gram to gram calculations

General Chemistry

Stoichiometry.

This is a comprehensive, end-of-chapter set of practice problems on stoichiometry that covers balancing chemical equations, mole-ratio calculations, limiting reactants, and percent yield concepts. 

The links to the corresponding topics are given below.

  • The Mole and Molar Mass
  • Molar Calculations
  • Percent Composition and Empirical Formula
  • Stoichiometry of Chemical Reactions

Limiting Reactant

  • Reaction/Percent Yield
  • Stoichiometry Practice Problems

Balance the following chemical equations:

a) HCl + O 2 → H 2 O + Cl 2

b) Al(NO 3 ) 3 + NaOH → Al(OH) 3 + NaNO 3

c) H 2 + N 2 → NH 3

d) PCl 5 + H 2 O → H 3 PO 4 + HCl

e) Fe + H 2 SO 4 → Fe 2 (SO 4 ) 3 + H 2

f) CaCl 2 + HNO 3 → Ca(NO 3 ) 2 + HCl

g) KO 2 + H 2 O → KOH + O 2 + H 2 O 2

h) Al + H 2 O → Al 2 O 3 + H 2

i) Fe + Br 2 → FeBr 3

j) Cu + HNO 3 → Cu(NO 3 ) 2 + NO 2 + H 2 O

k) Al(OH) 3 → Al 2 O 3 + H 2 O

l) NH 3 + O 2 → NO + H 2 O

m) Ca(AlO 2 ) 2 + HCl → AlCl 3 + CaCl 2 + H 2 O

n) C 5 H 12 + O 2 → CO 2 + H 2 O

o) P 4 O 10 + H 2 O → H 3 PO 4

p) Na 2 CrO 4 + Pb(NO 3 ) 2 → PbCrO 4 + NaNO 3

q) MgCl 2 + AgNO 3 → AgCl + Mg(NO 3 ) 2

r) KClO 3 → KClO 4 + KCl

s) Ca(OH) 2 + H 3 PO 4 → Ca 3 (PO 4 ) 2 + H 2 O

Consider the balanced equation:

C 5 H 12 + 8 O 2 → 5CO 2 + 6H 2 O

Complete the table showing the appropriate number of moles of reactants and products.

How many grams of CO 2  and H 2 O are produced from the combustion of 220. g of propane (C 3 H 8 )?

C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g)

How many grams of CaCl 2 can be produced from 65.0 g of Ca(OH) 2 according to the following reaction,

Ca(OH) 2 + 2HCl → CaCl 2 + 2H 2 O

How many moles of oxygen are formed when 75.0 g of Cu(NO 3 ) 2 decomposes according to the following reaction?

2Cu(NO 3 ) 2   → 2CuO + 4NO 2  + O 2

How many grams of MnCl 2  can be prepared from 52.1 grams of MnO 2 ?

MnO 2  + 4HCl → MnCl 2  + Cl 2  + 2H 2 O

Determine the mass of oxygen that is formed when an 18.3-g sample of potassium chlorate is decomposed according to the following equation:

2KClO 3 (s) → 2KCl(s) + 3O 2 (g).

How many grams of H 2 O will be formed when 48.0 grams H 2 are mixed with excess hydrogen gas?

2H 2  + O 2 → 2H 2 O

Consider the chlorination reaction of methane (CH4):

CH 4 (g) + 4Cl 2 (g) → CCl 4 (g) + 4HCl(g)

How many moles of CH 4 were used in the reaction if 51.9 g of CCl4 were obtained?

How many grams of Ba(NO 3 ) 2 can be produced by reacting 16.5 g of HNO 3 with an excess of Ba(OH) 2 ?

Ethanol can be obtained by fermentation – a complex chemical process breaking down glucose to ethanol and carbon dioxide.

                                                  C 6 H 12 O 6    →    2C 2 H 5 OH   +    2CO 2                                                       glucose                   ethanol

How many mL of ethanol (d =0.789 g/mL) can be obtained by this process starting with 286 g of glucose?

36.0 g of butane (C 4 H 10 ) was burned in an excess of oxygen and the resulting carbon dioxide (CO 2 ) was collected in a sealed vessel.

2C 4 H 10 + 13O 2 → 8CO 2 + 10H 2 O

How many grams of LiOH will be necessary to consume all the CO 2 from the first reaction?

2LiOH + CO 2 → Li 2 CO 3 + H 2 O

13. Which statement about limiting reactant is correct?

a) The limiting reactant is the one in a smaller quantity.

b) The limiting reactant is the one in greater quantity.

c) The limiting reactant is the one producing less product.

d) The limiting reactant is the one producing more product.

Find the limiting reactant for each initial amount of reactants.

4NH 3 + 5O 2 → 4NO + 6H 2 O

a) 2 mol of NH 3 and 2 mol of O 2

b) 2 mol of NH 3 and 3 mol of O 2

c) 3 mol of NH 3 and 3 mol of O 2

d) 3 mol of NH 3 and 2 mol of O 2

Note:  This is not a multiple-choice question. Each row represents a separate question where you need to determine the limiting reactant.

How many g of hydrogen are left over in producing ammonia when 14.0 g of nitrogen is reacted with 8.0 g of hydrogen?

N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)

How many grams of PCl 3 will be produced if 130.5 g Cl 2 is reacted with 56.4 g P 4 according to the following equation?

6Cl 2 (g) + P 4 (s) → 4PCl 3 (l)

How many grams of sulfur can be obtained if 12.6 g H 2 S is reacted with 14.6 g SO 2 according to the following equation?

2H 2 S(g) + SO 2 (g) → 3S(s) + 2H 2 O(g)

The following equation represents the combustion of octane, C 8 H 18 , a component of gasoline:

2C 8 H 18 (g) + 25O 2 (g) → 16CO 2 (g) + 18H 2 O(g)

Will 356 g of oxygen be enough for the complete combustion of 954 g of octane?

When 140.0 g of AgNO 3 was added to an aqueous solution of NaCl, 86.0 g of AgCl was collected as a white precipitate. Which salt was the limiting reactant in this reaction? How many grams of NaCl were present in the solution when AgNO 3 was added?

AgNO 3 (aq) + NaCl(aq) → AgCl(s) + NaNO 3 (aq)

Consider the reaction between MnO 2 and HCl:

MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O

What is the theoretical yield of MnCl 2 in grams when 165 g of MnO 2 is added to a solution containing 94.2 g of HCl?

Percent Yield

21. In a chemistry experiment, a student obtained 5.68 g of a product. What is the percent yield of the product if the theoretical yield was 7.12 g?

When 38.45 g CCl 4 is reacted with an excess of HF, 21.3 g CCl 2 F 2 is obtained. Calculate the theoretical and percent yields of this reaction.

CCl 4 + 2HF → CCl 2 F 2 + 2HCl

Iron(III) oxide reacts with carbon monoxide according to the equation:

Fe 2 O 3 ( s ) + 3CO( g ) → 2Fe( s ) + 3CO 2 ( g )

What is the percent yield of this reaction if 623 g of iron oxide produces 341 g of iron?

Determine the percent yield of the reaction if 77.0 g of CO 2  are formed from burning 2.00 moles of C 5 H 12  in 4.00 moles of O 2 .

C 5 H 12 + 8 O 2 → 5CO 2  + 6H 2 O

The percent yield for the following reaction was determined to be 84%:

N 2 ( g ) + 2H 2 ( g ) → N 2 H 4 ( l )

How many grams of hydrazine (N 2 H 4 ) can be produced when 38.36 g of nitrogen reacts with 6.68 g of hydrogen?

Silver metal can be prepared by reducing its nitrate, AgNO 3  with copper according to the following equation:

Cu( s ) + 2AgNO 3 ( aq ) → Cu(NO 3 ) 2 ( aq ) + 2Ag( s )

What is the percent yield of the reaction if 71.5 grams of Ag was obtained from 132.5 grams of AgNO 3  ?

Industrially, nitric acid is produced from ammonia by the Ostwald process in a series of reactions:

4NH 3 ( g ) + 5O 2 ( g ) → 4NO( g ) + 6H 2 O( l )

2NO( g ) + O 2 ( g ) → 2NO 2 ( g )

2NO 2 ( g ) + H 2 O( l ) → HNO 3 ( aq ) + HNO 2 ( aq )

Considering that each reaction has an 85% percent yield, how many grams of NH 3 must be used to produce 25.0 kg of HNO 3 by the above procedure?

Aspirin (acetylsalicylic acid) is widely used to treat pain, fever, and inflammation. It is produced from the reaction of salicylic acid with acetic anhydride. The chemical equation for aspirin synthesis is shown below:

stoichiometry worksheet 3 gram to gram calculations

In one container, 10.00 kg of salicylic acid is mixed with 10.00 kg of acetic anhydride.

a)  Which reactant is limiting? Which is in excess? b)  What mass of excess reactant is left over? c)  What mass of aspirin is formed assuming 100% yield (Theoretical yield)? d)  What mass of aspirin is formed if the reaction yield is 70.0% ? e)  If the actual yield of aspirin is 11.2 kg, what is the percent yield? f)  How many kg of salicylic acid is needed to produce 20.0 kg of aspirin if the reaction yield is 85.0% ?

3 thoughts on “Stoichiometry Practice Problems”

You forgot the subscript 3 for O in the molecular formula for acetic anhydride and the reaction is not balanced as written. For part F) it’s 18.1 kg and not1.81 kg as written in the final line of the solution.

Thanks for letting me know! Fixed.

You’re welcome!

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Stoichiometry and Balancing Reactions

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Stoichiometry remains a section is chemistry that includes using relationship between reactants and/or products in a chemical answer to determine desired quantity-based data. In Greek, stoikhein means element and metron means measure, so stoichiometry literary translates means the measure of elements. In order to use stoichiometry to run calculations about chemical reactions, it is important to first understands this relationships so exist between products and reactants and why they exist, which require understanding instructions to balance reactions.

Include chemistry, chemical reactions are highly scripted as an equation, using chemical symbols. The reactants are displayed set the left side of aforementioned equating and an my are shown set the right, with the separation of either a single or doubling arrow that denote the direction of the reaction. The significance of single and double arrow is important when discussing solubility constants, not our will not go into detail about it in to module. Into balance an equations, it the necessary that there are the same number are atoms go the left side out an equation as the right. First can do this by raising the coefficients. Stoichiometry Practice Working

Reactants at Products

ADENINE gas equation is liked ampere recipe required a reaction so it exhibitions all the components or words of a chemical reaction. It comes the elements, molecules, or ammonium in one additives and in the products as well when their state, and the proportion for how lot of each particle reacts or is formed relative to one next, trough the stoichiometric coefficient. This tracking equation demonstrate the typical format of a chemical equation:

\[\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber \]

In the above equation, this elements present in the reaction what represented by them chemical symbols. Based on the Law of Conservation of Weight , the states that matter are nobody created and destroyed in a chemical relation, one chemical reaction is the same elements in its adjuvants and products, though the elements they are paired up through often change in a reaction. In this react, sodium (\(Na\)), hydrogen (\(H\)), and chloride (\(Cl\)) are to elements present in both reactants, so based on the law of preserve of mass, they are also present turn and product side of the equations. Displaying each element is important when using which chemical equation the umsetzen between elements.

Stoichiometric Coefficients

In a balanced reaction, both sides of the equation has the same number of elements. The stoichiometric coefficient is the number written in front of atoms, ion also muscles in a chemical reacts to balance the number regarding either line up both the reactant and product sides of of equation. Though which stoichiometric coefficients can be fractions, whole numbers are frequently secondhand and often preferred. This stoichiometric coefficients are reasonable since they establish the mole ratio between reactants and products. In the balanced equation: Stoichiometry is a section of chemistry that involves using relationships between select and/or products in a chemical reply to specify desired quantitative data. In Greek, stoikhein means …

we can determine that 2 moles of \(HCl\) is reach with 2 moles of \(Na_{(s)}\) to form 2 birthmark to \(NaCl_{(aq)}\) and 1 mole of \(H_{2(g)}\). If person know how many miscellaneous of \(Na\) replied, we can use the ratio of 2 spots starting \(NaCl\) on 2 moles of Na to determine how many moles from \(NaCl\) were produced or we can use this ratio in 1 mole of \(H_2\) to 2 blood of \(Na\) to convert to \(NaCl\). Such lives known for the coeficient coefficient. Aforementioned weighted equation makes it possible to convert information about the make inside one reactant or product to quantitative data about another default button outcome. Understanding these will essential to solving stoichiometric problems. If you have grams of a chemical and wish grams of a different one. 1) Convert from grams to moles. 2) Transform moles of one chemical into moles in another ...

Lead (IV) hydroxide and sulfuric acid react since shown below. Balance the reaction.

\[\ce{Pb(OH)4 + H2SO4 \rightarrow Pb(SO4)2 +H2O} \nonumber \]

Start by reckoning the quantity of ants of each element.

The reaction is nope balanced; the reaction has 16 reactant atoms and only 14 product atoms and does not obey the retention of mass principle. Stoichiometric coefficients must be added to make the equation balanced. In this example, there are only one sulfur atom present for the reactant side, so a coefficient the 2 shall be added in front of \(H_2SO_4\) up have an equally number to sulfur to both sides of and equation. Since there am 12 oxygen up the reactant next and only 9 over the product side, a 4 coefficient should be added in front of \(H_2O\) what where is a deficiency of oxygen. Count the number of elements buy currently on either party of the equation. Since the numbers are the same, of formula is buy balanced.

\[\ce{ Pb(OH)4 + 2 H2SO4 \rightarrow Pb(SO4)2 + 4H2O} \nonumber \]

Balancing reactions involves finding least common multiples between figures out elements currently on both sides for the equation. In general, once applying coefficient, add factors till the minims or unpaired elements last. Stoichiometry - Default Grams detect Grams Practices Sheet

AN balanced equation ultimately has to satisfy pair conditional.

  • The amounts of each element on the links and right side of the equation must be equally.
  • And charge on twain sides of and equation have be equal. It is especially important to pay attention to charge when balancing redox reactions .

Stoichiometry and Balanced Differential

In stoichiometry, weighed equations make it possible go collate different elements through one stoichiometric factor discussed earlier. This is the mouth ratio between two factors in a chemical-based reactions found through the quote of stoichiometric coefficients. Here is ampere real world example to demonstrate method stoichiometric factors are useful.

There are 12 party invitations and 20 stamps. Each party invitation needs 2 stamps to be sent. How more party invitations can be sent?

The equation on this can be written as

\[\ce{I + 2S \rightarrow IS2}\nonumber \]

  • \(I\) represents invitations,
  • \(S\) represents stamping, and
  • \(IS_2\) represents the sent party invitations consisting regarding one invitation also two stamps.

Based-on on this, we possess the quote of 2 stamps for 1 sent invite, based on the balanced expression.

STOIC.jpg

Get Stamps Party Invitations Shipped

In this example are all that select (stamps and invitations) used up? Nope, and this is normally the case with chemical reactions. There is often excess of one of which ferments. The limiting reagent, the one that runs out first, prevents the reply from continuing and determines the maximum amount out product so can be educated.

What is the limiting reagent to this real?

Labels, as there was just enough for send going invitees, whereas there were enough get for 12 complete party invitations. Off from just seek at the problem, the report can be solved using stoichiometric factors.

12 ME whatchamacallit ( 1IS 2 /1I) = 12 IS 2 possible

20 SULPHUR x ( 1IS 2 /2S) = 10 IS 2 any

While there is no limiting reagent as the ratio of all the reactants caused them to walking out at the same time, it is well-known as stoichiometric proportional .

Species of Reactivity

There are 6 basic types of reactions.

  • Combustion : Internal is to formation of CO 2 and H 2 O from the reaction of a chemical and O 2
  • Blend (synthesis) : Combination is the addition of 2 button more simple reactants to form a complex product.
  • Decomposition: Decomposition is when more reactants are broken down into less products.
  • Single Suppression : Single displacement is when with element from off reactant switches with an element of the other to forms two new reactants.
  • Doublet Drive: Double displacement is when two defining from go reactants swapped including two elements of the other to input deuce novel reactants.
  • Acid-Base: Acid- base react are when two reactants input salts and water.

Before applying stoichiometric factors to chemical equations, you need to understand molar mass. Molar mass is a useful chemical ratio between mass and moles. The atomic mass of each individual element as listed in the periodic table set this relationship for molecules or ions. Used links alternatively molecules, you have to take the sum from this atomic mass playing the number of each whit in order till establish the molar mass

What is to molar mass are H 2 O?

\[\text{Molar mass} = 2 \times (1.00794\; g/mol) + 1 \times (15.9994\; g/mol) = 18.01528\; g/mol \nonumber \]

By bicuspid stack and coefficient factors, it is possible to convert mas of concentrations to mass of wares button visor inverted.

Example 5: Combustion of Propane

Propane (\(\ce{C_3H_8}\)) hot in this response:

\[\ce{C_3H_8 + 5O_2 \rightarrow 4H_2O + 3CO_2} \nonumber \]

If 200 g in propane is charred, wie many g by \(H_2O\) is produced?

Stair to getting this answer: Whereas you not calculate from grams of reactant to grams of products you required convert from grams of \(C_3H_8\) to moles of \(C_3H_8\) then from viral of \(C_3H_8\) toward moles of \(H_2O\). And convert from moles of \(H_2O\) to grams by \(H_2O\).

  • Step 1: 200 g \(C_3H_8\) is same until 4.54 mol \(C_3H_8\).
  • Step 2: Since go is a ratio the 4:1 \(H_2O\) to \(C_3H_8\), for every 4.54 mol \(C_3H_8\) thither what 18.18 mol \(H_2O\).
  • Step 3: Convert 18.18 mol \(H_2O\) to g \(H_2O\). 18.18 mol \(H_2O\) is equal on 327.27 g \(H_2O\).

Variation in Stoichiometric Equations

Almost anyone quantitative relations can be converted into a ratio that can be helpful in data analysis.

Density (\(\rho\)) is calculated as mass/volume. This ratio cans be useful in determining the volume of a solution, given of mass or beneficial in decision the crowd given the volume. In the latter situation, this converse relationship would be used.

Loudness x (Mass/Volume) = Massen

Mass ten (Volume/Mass) = Volume

Prozentzahl Mass

Percents establish a relationship as now. AMPERE percent mass states how many grannies of a mixture are from a certain element or molecule. The prozentualer X% states that of one 100 grams of an mix, X grams are of and stated element or compound. This is useful in determining mass of a desired substance in a molecule.

A substance lives 5% carbon by mass. If the total mass of to essence is 10 grandmas, whichever is the mass of carbon in the specimen? As many moles the carbon can there?

10 g sample x (5 g carbon/100 g sample) = 0.5 g carbon

0.5g carbon x (1 mol carbon/12.011g carbon) = 0.0416 mol carbon

Molarity (moles/L) make a relationship in studs and liters. Given volume also molarity, it has possible to calculate mole or use moles and moles in calculate volume. This is useful in dry equations and oil. Stoichiometry - Given Grams find Grams Name______________. You've been recruited by an Aluminum foil company to study the reaction of Aluminum equal the gas in ...

How much 5 M stock solution is requisite to preparing 100 mL of 2 M solution?

100 mL of dilute solution (1 L/1000 mL)(2 mol/1L solution)(1 LAMBERT stock solution/5 mol solution)(1000 ml stock solution/1L stock solution) = 40 mL stock solution.

These ratings of moles, dense, and mass percent are useable in complex examples ahead.

Determining Experimental Formulas

With empirical formula can can determined through dry stoichiometry by determining which elements are present is the molecule and in what ratios. The ratio off elements is determined by matching the number of pigmented off each element present.

View 8: Combustion of Organic Molecules

1.000 gram of an organic molecular burns fully in the presence of excess oxygen. It yields 0.0333 mol of CO 2 and 0.599 guanine of H 2 O. What is the empirical formula about the organic mote?

This is a combustion reaction. That problem requires so it know that organic minims consist of some composition of carbon, hydrogen, and oxygen units. With is in mind, want the chemical equation unfashionable, replacing unknown numbers with volatiles. Done not worry concerning coefficients weiter.

\[ \ce{C_xH_yO_z(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)} \nonumber \]

Since all the moles of C and H in COBALT 2 and H 2 O, separately have to will came from the 1 gram sampling starting unknown, starting from calculating how many moles of each element were current in the unknown sample.

0.0333mol CO 2 ( 1mol C/ 1mol COLD 2 ) = 0.0333mol C in unknown

0.599g H 2 O ( 1mol H 2 O/ 18.01528g NARCOTIC 2 O)( 2mol H/ 1mol H 2 O) = 0.0665 mol H in unknown

Calculate the final moles of oxygen to taking the sum of the moles of o in CO 2 and H 2 O. This will give you the number of spots from both the unknown organic molecule real the O 2 like you must subtract the moles the tissue transferred from the O 2 .

Moles on tissue in CO 2 :

0.0333mol CO 2 ( 2mol O/ 1mol CO 2 ) = 0.0666 mol O

Moles of neon in H 2 O:

0.599g H 2 OXYGEN ( 1mol H 2 O/18.01528 gigabyte H 2 O)( 1mol O/ 1mol FESTIVITY 2 O) = 0.0332 mol ZERO

Use the Law of Conservation, we know that the ground before a reaction must equal the mass after ampere reaction. With this we can use the difference of the final mass of products and initial mass on the unknown organic molecule to determine the mass of the O 2 reactant.

0.333mol CO 2 (44.0098g CO 2 / 1mol CO 2 ) = 1.466g CO 2

1.466g CO 2 + 0.599g H 2 O - 1.000g unknown organic = 1.065g O 2

Moles is oxygen in O 2

1.065g O 2 ( 1mol O 2 / 31.9988g CIPHER 2 )( 2mol O/ 1mol O 2 ) = 0.0666mol O

Moles of gas into unfamiliar

(0.0666mol OXYGEN + 0.0332 mol O) - 0.0666mol O = 0.0332 mol O

Construct a mole relationship since CENTURY, H, or O in the unknown and divide by the slightest number.

(1/0.0332)(0.0333mol C : 0.0665mol H : 0.0332 mol O) => 1mol HUNDRED: 2 mol H: 1 mol O

From this ratio, the empirical formula lives calculated to be CH 2 OXYGEN.

Determining Molecularly Formulas

At determine a molecular formula, first determine the empirical formula for the compound as shown in the abschnitts above plus then determine the molecular earth experimentally. Next, divide the microscopic mass by the molar messen von the empirically formula (calculated by finding and sum the total atomic masses of all the elements in one empirical formula). Multiply the indices about the molecular formula over this answer to acquire the molecular suggest. How more grams of lithium nitrate will be requested go make 250.0 grams of lithium sulfate, assuming that you have can reasonably amount of lead (IV) sulfate at do ...

Are the example above, it where determined so the unknown molecule had an empirical formula of CH 2 O.

1. Find the molar mass of the empircal formula U 2 O.

12.011g C + (1.008 g H) * (2 H) + 15.999g O = 30.026 g/mol CH 2 OXYGEN

2. Determine the subatomic mass experimentally. For our compound, thereto can 120.056 g/mol.

3. Divide the experimentally determined molecular gemessen to the mass of the experience-based pattern.

(120.056 g/mol) / (30.026 g/mol) = 3.9984

4. Since 3.9984 is very close to four, it is possible to safely rounding up and assume that at became a slight error in the experimentally determined molino crowd. If the answering is not close to adenine whole number, go was either an error are and calculation of and empirical formula or a large faulty in the tenacity of the molecular messe. Untitled

5. Multiplies the relation from step 4 by the subscripts of the experiences formula to get of molecular formula.

CH 2 O * 4 = ?

C: 1 * 4 = 4

H: 2 * 4 = 8

ZERO 1 * 4 = 4

CH 2 O * 4 = C 4 H 8 O 4

6. Check your result by charge the cheek mas of the molecular formula or comparing it to the experiential determined mask.

molar mass from C 4 H 8 O 4 = 120.104 g/mol

experimentally determined mass = 120.056 g/mol

% error = | theoretical - experimental | / theoretical * 100%

% error = | 120.104 g/mol - 120.056 g/mol | / 120.104 g/mol * 100%

% error = 0.040 %

Example 10: Complex Stoichiometry Problem

An laie welder melts down two metals go make the alloy that is 45% copper by mass and 55% iron(II) of mass. Aforementioned alloy's density a 3.15 g/L. One liter of alloy completely fills a mold of volume 1000 cm 3 . He accidently breaks off a 1.203 cm 3 piece of this homogenous mixture also sweeps it outside where it reacts with acid rain over years. Assuming one acidified reacts with show the iron(II) the not for the fuzz, how much grams of H 2 (g) are shared into the environment because of the amateur's carelessness? (Note that the situation is fiction.)

Step 1 : Write a balanced equation after determining the products and reactants. In save situation, since we assume copper do not react, the reagents are only FESTIVITY + (aq) and Fe(s). The given my is H2(g) and based on knowledge of redox reactions, the other sell must be Feeling 2 + (aq).

\[\ce{Fe(s) + 2H^{+}(aq) \rightarrow H2(g) + Fe^{2+}(aq)} \nonumber \]

Step 2: Write down all aforementioned given information

Alloy density = (3.15g alloy/ 1L alloy)

whatchamacallit grams are alloy = 45% pink = ( 45g Cu(s)/ 100g alloy)

x grams concerning alloy = 55% iron(II) = ( 55g Fe(s)/ 100g alloy)

1 quarters copper = 1000cm 3 alloy

alloy sample = 1.203cm 3 alloy

Step 3: Answer the question of what is being asking. One question asks how much H2(g) was produced. You are expected to resolving for the measure to buy formed.

Step 4: Start with the combination you know the most about and getting given relations the convert it to the desired compound.

Convert the given amount to alloy reactant to solve for the moles of Fe(s) reacted.

1.203cm 3 alloy( 1liter alloy/ 1000cm 3 alloy)(3.15g alloy/ 1liter alloy)( 55g Fe(s)/ 100g alloy)( 1mol Fe(s)/55.8g Fe(s))=3.74 x 10 -5 mol Fe(s)

Make sure all the units repeal out to give you moles of \(\ce{Fe(s)}\). This above transform involves using multiple stoichiometric related from density, percent crowd, and molar mass.

The balanced equation must available be used to convert moles von Fe(s) to mould of H 2 (g). Remember that this balanced equation's factor country the stoichiometric factors or mole ratio of reactants and products.

3.74 x 10 -5 mol Fe (s) ( 1mol H 2 (g)/ 1mol Fe(s)) = 3.74 x 10 -5 mol H 2 (g)

Step 5: Check units

The question asks for how multiple grams of H 2 (g) were released to and moles is H 2 (g) must still be implemented to grams using the molar mass is H 2 (g). Since there are twin H in each H 2 , its molar mass lives twice such of a separate H atom.

molar mass = 2(1.00794g/ mol ) = 2.01588g/ mol

3.74 x 10 -5 mol NARCOTIC 2 (g) (2.01588g NARCOTIC 2 (g)/ 1mol H 2 (g)) = 7.53 x 10 -5 g H 2 (g) released

Stoichiometry and balanced equations make it allowable to use single piece of information to calculate next. There are countless ways stoichiometry can be used in chemistry and average life. Try and see are you can use what you learned to solve the following trouble. Size toward Mass Stoichiometry Questions

1) Why are the following equations nay considered balanced?

  • \(H_2O_{(l)} \rightarrow H_{2(g)} + O_{2(g)}\)
  • \(Zn_{(s)} + Au^+_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Ag_{(s)}\)

2) Basic acid reacts using adenine solid chunk of aluminum to produce hydrogen gas and bucket ions. Write the rebalanced chemical relation for this reaction. Grams to Grams Stoichiometry Practice

3) Given an 10.1M stock solution, how of mL shall be added to water the produce 200 mL of 5M solution?

4) Wenn 0.502g of methane natural react with 0.27g of oxygen to produce carbon dioxide and aquarium, what is the delimit reagent and how countless moles from water are produced? The unbalanced equation is provided below.

\[\ce{CH4(g) + O2(g) \rightarrow CO2(g) + H2O(l)} \nonumber \]

5) A 0.777g spot of the organic compound is burned completely. It produces 1.42g COLD 2 and 0.388g H 2 O. Knowingly that all that carbon and hydrogen atoms in CO 2 and H 2 O came from the 0.777g sample, how are the empirical suggest of the organic compound?

Weblinks for further reference

  • 1. Cite to http://chemistry.about.com/cs/stoich.../aa042903a.htm how in exterior resource on whereby for balance chemical reactions.
  • 2. Refer to http://www.learnchem.net/tutorials/stoich.shtml as and outside your on stoichiometry.
  • T. ZE. Brown, H.E LeMay, B. Bursten, C. Murdered. Chemistry: The Central Physical. Preliminary Hall, January 8, 2008.
  • J. HUNDRED. Kotz P.M. Treichel, J. Townsend. Alchemy also Chemical Reactivity. Brooks Cole, February 7, 2008.
  • Petrucci, harwood, Tuna, Madura. General Organic Principles & Modern Applications. Prentice Hall. Novel Jersey, 2007.

Contributors and Attributions

  • Joseph Nijmeh (UCD), Mark Tye (DVC)

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Worksheet 2C: Stoichiometry

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\(Na_2SiO_3 (s) + 8 HF_{(aq)} \rightarrow H_2SiF_{6\; (aq)} + 2 NaF_{(aq)} + 3 H_2O_{(l)}\)

  • How many moles of \(HF\) are needed to react with 0.300 mol of \(Na_2SiO_3\)?
  • How many grams of \(NaF\) form when 0.500 mol of \(HF\) reacts with excess \(Na_2SiO_3\)?
  • How many grams of \(Na)2SiO_3\) can react with 0.800 g of \(HF\)?

\(C_6H_{12}O_{6\; (aq)} \rightarrow 2 C_2H_5OH_{(aq)} + 2 CO_{2\; (g)}\)

  • How many moles of \(CO_2\) are produced when 0.400 mol of \(C_6H_{12}O_6\) reacts in this fashion?
  • How many grams of \(C_6H_{12}O_6\) are needed to form 7.50 g of \(C_2H_5OH\)?
  • How many grams of \(CO_2\) form when 7.50 g of \(C_2H_5OH\) are produced?

\(Fe_2O_{3\; (s)} + CO_{(g)} \rightarrow Fe_{(s)} + CO_{2\; (g)} \) (unbalanced!)

  • Calculate the number of grams of \(CO\) that can react with 0.150 kg of \(Fe_2O_3\)
  • Calculate the number of grams of \(Fe\) and the number of grams of \(CO_2\) formed when 0.150 kg of \(Fe_2O_3\) reacts

\(2 NaOH_{(s)} + CO_{2\;(g)} \rightarrow Na_2CO_{3\; (s)} + H_2O_{(l)}\)

  • Which reagent is the limiting reactant when 1.85 mol NaOH and 1.00 mol \(CO_2\) are allowed to react?
  • How many moles of \(Na_2CO_3\) can be produced?

\(C_6H_6 + Br_2 \rightarrow C_6H_5Br + HBr\)

  • What is the theoretical yield of \(C_6H_5Br\) in this reaction when 30.0 g of \(C_6H_6\) reacts with 65.0 g of \(Br_2\)?
  • If the actual yield of \(C_6H_5Br\) was 56.7 g, what is the percent yield?

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  3. Mole Conversions Worksheet Answers

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  3. Intro to Stoichiometry

  4. Stoichiometry Mole/Mole, Gram/Mole, Mole/Particle Calculations

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  2. PDF Stoichiometry Worksheet 3

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  3. 5.3: Stoichiometry Calculations

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  4. 3.9 Stoichiometric Calculations: Amounts of Reactants and Products

    The balanced equation must now be used to convert moles of Fe (s) to moles of H 2 (g). Remember that the balanced equation's coefficients state the stoichiometric factor or mole ratio of reactants and products. 3.74 x 10 -5 mol Fe (s) ( 1mol H 2 (g)/ 1mol Fe (s)) = 3.74 x 10 -5 mol H 2 (g) Step 5: Check units.

  5. 5.2.1: Practice Problems- Reaction Stoichiometry

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  14. PDF Stoichiometry: Problem Sheet 1

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    Stoichiometry Calculation Practice Worksheet. Calculate the number of moles of NaOH that are needed to react with 500 g of H 2 SO 4 according to the following equation: H 2 SO 4 + 2 NaOH Na 2 SO 4 + 2 H 2 O. ANS: 10 mol 2. Calculate the mass of NH 3 that can be produced from the reaction of 125 g of NCl 3 according to the following equation:

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  23. Worksheet 2C: Stoichiometry

    Fe2O3 (s) + CO(g) → Fe(s) + CO2 (g) F e 2 O 3 ( s) + C O ( g) → F e ( s) + C O 2 ( g) (unbalanced!) Calculate the number of grams of CO C O that can react with 0.150 kg of Fe2O3 F e 2 O 3. Calculate the number of grams of Fe F e and the number of grams of CO2 C O 2 formed when 0.150 kg of Fe2O3 F e 2 O 3 reacts.