Stoichiometry Worksheet

Part 1

1. Use the balanced chemical reaction for the combustion of propane, C₃H₈, to answer the questions below.

C₃H₈ + 5 O₂ à 3 CO₂ + 4 H₂O

A) Write the equivalence statements based on the balanced chemical reaction.

a) ____1_____C₃H₈ = ____5_____ O₂ ______1______C₃H₈ = _____3____ CO₂

______1_____C₃H₈ = _____4____ H₂O

b) _______5____O₂ = ___1_______ C₃H₈ _____5______O₂ = ______3____ CO₂

_____5______O₂ = ____4______ H₂O

c) How many molecules of carbon dioxide will be produced if 20 molecules of propane react?

? cule CO₂ =	20 C₃H₈	3 CO₂
? cule CO₂ =	1	1 C₃H₈

? cule CO₂ = 60 CO₂ molecules

d) How many molecules of oxygen react with 20 molecules of propane?

? cule O₂ =	20 C₃H₈	5 O₂
? cule O₂ =	1	1 C₃H₈

? cule O₂ = 100 O₂ molecules

e) If 100 molecules of oxygen react, how many molecules of water will be produced?

? cule H₂O =	100 O₂	4 H₂O
? cule H₂O =	1	5 O₂

? cule H₂O = 80 H₂O molecules

f) If 3.75 E 23 molecules of oxygen react, how many molecules of carbon dioxide will be produced?

? cule CO₂ =	3.75 E23 O₂	3 CO₂
? cule CO₂ =	1	5 O₂

? cule CO₂ = 2.25 E23 CO₂ molecules

g) How many moles of carbon dioxide will be produced if 20.0 moles of oxygen react?

? mol CO₂ =	20.0 mol O₂	3 CO₂
? mol CO₂ =	1	5 O₂

? cule CO₂ = 12.0 mol CO₂

h) If 8.50 moles of water is produced by this reaction, how many moles of oxygen reacted?

? mol O₂ =	8.50 mol H₂O	5 O₂
? mol O₂ =	1	4 H₂O

? cule O₂ = 10.6 mol O₂

2. Use the balanced chemical reaction for the combustion of propane, C₃H₈, to answer the questions below.

C₃H₈ + 5 O₂ à 3 CO₂ + 4 H₂O

A) A student conducts an experiment using the reaction above. When she burns 44.0 grams of propane, 160.0 grams of oxygen is consumed. 132 grams of carbon dioxide and 72 grams of water are produced. Does this data support the Conservation of Mass? Use the masses given to support both of your answers.

Yes this data supports the conservation of mass. Number of atoms in reactants equals the number of atoms in products—atoms are neither created nor destroyed. The atoms are just getting rearranged. The data shows this because the total mass of the reactants is 204.0 grams and the total mass of the products is 204.0 grams.

How are these masses related to the coefficients in the balanced reaction? Use the masses given to support both of your answers.

The ratios of the coefficients in a balanced reaction do work for mass. This means you must use the mass to find the moles of each substance and then see if the mole ratios are the same as the coefficient ratios in the balanced reaction.

? mole propane = (44.0 g propane/1) x ( 1 mol/44.0 g) = 1.00 mole

? mole oxygen = (160.g/1) x ( 1 mol/ 32.0 g) = 5.00 mole

? mole carbon dioxide = (132 g/ 1) x (1 mol / 44.0 g) = 3.00 mole

? mole water = (72.0 g/ 1) x (1 mol / 18.0 g) = 4.00 mole

Notice the ratio of moles is the same as the ratio of coefficients.

B) Write the equivalence statements based on the balanced chemical reaction.

a) _________1___C₃H₈ = ____5______ O₂ _____1_______C₃H₈ = _3_________ CO₂

__1__________C₃H₈ = _______4___ H₂O

b) _________5___O₂ = ______1_____ C₃H₈ _______5_____O₂ = ______3_____ CO₂

_______5_____O₂ = ______4_____ H₂O

B) How many moles of O₂ will react with 2.00 moles of propane?

? mol O₂ =	2.00 mol C₃H₈	5 O₂
? mol O₂ =	1	1 C₃H₈

? mol O₂ = 10.0 mol O₂

C) How many moles of carbon dioxide will be produced if 2.00 grams of O₂ react?

? mol CO₂ =	2.00 g O₂	1 mol O₂	3 CO₂
? mol CO₂ =	1	32.00 g O₂	5 O₂

? cule CO₂ = .0375 mol CO₂

D) How many grams of propane react to produce 8.00 grams of water?

? grams C₃H₈ =	8.00 g H₂O	1 mol H₂O	1 C₃H₈	44.11 g C₃H₈
? grams C₃H₈ =	1	18.02 g H₂O	4 H₂O	1 mol C₃H₈

? gram C₃H₈ = 4.90 grams C₃H₈

3. Laughing gas, N₂O can be turned into smog NO₂ by heating the laughing gas in the presence of oxygen.

2N₂O (g) + 3O₂ (g) à 4NO₂ (g)

A) 9.00 grams of laughing gas react. How many moles of oxygen react?

? mol O₂ =	9.00 g N₂O	1 mol N₂O	3 O₂
? mol O₂ =	1	44.00 g N₂O	2 N₂O

? mol O₂ = .307 mol O₂

B) If 7.50 grams of oxygen react, how many grams of smog are produced?

? g NO₂ =	7.50 g O₂	1 mol O₂	4 NO₂	46.0 g NO₂
? g NO₂ =	1	32.00 g O₂	3 O₂	1 mol NO₂

? g NO₂ = 14.4 g NO₂

4. The acid in your stomach that aids in breaking down proteins is called hydrochloric acid, HCl. Occasionally the glands that produce the HCl make more than is needed, and you get those nasty sour burbs. You can neutralize the excess acid by taking milk of magnesia. The active ingredient in milk of magnesia is Mg(OH)₂. The chemical reaction is shown below.

2 HCl + Mg(OH)₂ à MgCl₂ + 2H₂O

A) If you drink 2.00 grams of Mg(OH)₂, how many moles of HCl are neutralized?

? mol HCl =	2.00 g Mg(OH)₂	1 mol Mg(OH)₂	2 HCl
? mol HCl =	1	58.345 g Mg(OH)₂	1 Mg(OH)₂

? mol HCl = .0689 mol HCl

B) If your glands produced 8.33 x 10²¹molecules, how many grams of Mg(OH)₂ must you drink to neutralize all of this acid?

? g Mg(OH)₂ =	8.33 E21 HCl	1 mol HCl	1 Mg(OH)₂	58.345 g Mg(OH)₂
? g Mg(OH)₂ =	1	6.022 E23 HCl	2 HCl	1 mol Mg(OH)₂

? g Mg(OH)₂ = .404 g Mg(OH)₂

5. Consider the double replacement reaction below. The lead(II) idodide is a bright yellow insoluble substance. Prior to the 1960’s it was used as a dye in yellow paint. Lead poisoning could result if you eat the paint so is different dye is used in today’s paint.

Sodium iodide + lead (II) nitrate à Sodium nitrate + Lead (II) iodide

A) Write a balanced reaction.

2NaI + Pb(NO₃)₂à 2 NaNO₃ + PbI₂

B) If 10.0 grams of sodium iodide react, how many grams of lead (II) iodide are produced.

? g PbI₂ =	10.0 g NaI	1 mol NaI	1 PbI₂	461.00 g PbI₂
? g PbI₂ =	1	149.9 g NaI	2 NaI	1 mol PbI₂

? g PbI₂ = 15.4 g PbI₂

C) When a student completes this reaction she filters and dries the lead (II) iodide. The mass of the yellow solid is 8.67 grams. What is the actual, theoretical and percent yield?

The actual yield is just what she obtained in the experiment. It is 8.67 grams.

The theoretical yield is what stoichiometry predicts. It is 15.4 grams. (See part B above for calculation.)

The percentage yield compares the actual yield to the theoretical yield.

% Yield = (Actual Yield/Theoretical Yield) x 100

% Yield = (8.67 g / 15.4 g) x 100 = 56.4 % She lost a lot of the product during the experiment.

Part 3 Limiting Reactants

6. Thionyl chloride, SOCl₂, is used as a powerful drying agent. The thionyl chloride reacts with water as shown by the reaction below.

SOCl₂ (l) + H₂O (l) à SO₂ (g) + 2HCl (g)

A) 35.0 grams of thionyl chloride is placed in a glass containing 500.0 grams of water. Which substance is the limiting reactant? Briefly explain your answer.

First find the moles of each substance.

? mol SOCl₂=	35.0 g SOCl₂	1 mol SOCl₂
? mol SOCl₂=	1	118.96 g SOCl₂

? mol SOCl₂= .294 mol SOCl₂

? mol H₂O =	500.0 g H₂O	1 mol H₂O
? mol H₂O =	1	18.02 g H₂O

? mol H₂O = 27.8 mol H₂O

There are several ways to find which is the limiting reactant. My favorite method is to find the ratio of the mole : coefficient in the balanced reaction. Whichever substance had the smaller ratio is the limiting reactant.

294 mole /1=.294à limiting! B/c smaller ratio

27.8mole/1=27.8àexcess b/c bigger ratio!

Notice the comparison of moles/coefficient leads to the substance that is the limiting reactant, but this ratio is never used in the calculations.

B) 35.0 grams of thionyl chloride is placed in a glass containing 500.0 grams of water. How many grams of water will react? How many grams of thionyl chloride will react?

? g H₂O =	35.0 g SOCl₂	1 mol SOCl₂	1 H₂O	18.02 g H₂O
? g H₂O =	1	118.96 g SOCl₂	1 SOCl₂	1 mol H₂O

? g H₂O = 5.30 g H₂O react

All 35.0 grams of SOCl₂ react as it is the limiting reactant.

C) 35.0 grams of thionyl chloride is placed in a glass containing 500.0 grams of water. After the reaction is complete how many grams of water will remain unreacted? How many grams of the thionyl chloride will remain unreacted?

If you start will 500.0 grams of water and 5.30 grams of water react then you must subtract to find the mass of water that remain after the reaction is complete.

494.7 grams of water remain unreacted.

Since thionyl chloride is the limiting reactant, all of it reacts to make the product.

D) 35.0 grams of thionyl chloride is placed in a glass 500 grams of water. How many HCl molecules will be produced by this reaction?

Use limiting to determine all others!!!

? cule HCl =	35.0 g SOCl₂	1 mol SOCl₂	2 HCl	6.022 E23 HCl
? cule HCl =	1	118.96 g SOCl₂	1 SOCl₂	1 mol HCl

? molecules HCl = 3.54 E23 HCl cules
7. For each of the unbalanced reactions below5.00 grams of each reactant is mixed together.

A) Balance the reaction.

B) Determine which substance is the limiting reactant.

C) Determine the mass of the excess substance that remains after the reaction is complete.

D) Find the mass of one of the products made by the reaction.

System #1

CaC₂ + 2H₂O à Ca(OH)₂ + C₂H₂

First balance the reaction.

Which substance is the limiting reactant? Briefly explain.

Determine which substance is the limiting reactant. Find the moles of each substance. Next find the ratio of moles : coefficient. The smallest ratio is the limiting reactant.

? mol CaC₂ =	5.00 g CaC₂	1 mol CaC₂
? mol CaC₂ =	1	64.10 g CaC₂

? mol CaC₂ = .0780 mol CaC₂

? mol H₂O =	5.00 g H₂O	1 mol H₂O
? mol H₂O =	1	18.02 g H₂O

? mol H₂O = .278 mol H₂O

.0780/1àlimiting reactant is CaC₂ !

.278/2=.138 -> excess is H₂O

Determine the mass of excess that remains unreacted. Show your work.

First find the mass of water that reacts. Then subtract the mass that reacts from 5.00 grams to get the amount of water that remains unreacted.

? g H₂O =	5.00 g CaC₂	1 mol CaC₂	2 H₂O	18.02 g H₂O
? g H₂O =	1	64.10 g CaC₂	1 CaC₂	1 mol H₂O

? g H₂O = 2.81 g H₂O react

5.00-2.81=2.19g excess

What mass of Ca(OH)₂ is produced in the reaction? Show your work.

Use the limiting reactant to find the mass of product made during the reaction.

? g Ca(OH)₂ =	5.00 g CaC₂	1 mol CaC₂	1 Ca(OH)₂	74.10g Ca(OH)₂
? g Ca(OH)₂ =	1	64.10 g CaC₂	1 CaC₂	1 mol Ca(OH)₂

? g Ca(OH)₂ = 5.78 g Ca(OH)₂

System #2

Na₂B₄O₇ + H₂SO₄ + 5H₂O à 4H₃BO₃ + Na₂SO₄

First balance the reaction.

Which substance is the limiting reactant? Briefly explain.

Determine which substance is the limiting reactant. Find the moles of each substance. Next find the ratio of moles : coefficient. The smallest ratio is the limiting reactant.

? mol Na₂B₄O₇ =	5.00 g Na₂B₄O₇	1 mol Na₂B₄O₇
? mol Na₂B₄O₇ =	1	201.24 g Na₂B₄O₇

? mol Na₂B₄O₇ = .0248 mol Na₂B₄O₇

? mol H₂SO₄ =	5.00 g H₂SO₄	1 mol H₂SO₄
? mol H₂SO₄ =	1	98.08 g H₂SO₄

? mol H₂SO₄ = .0510 mol H₂SO₄

? mol H₂O =	5.00 g H₂O	1 mol H₂O
? mol H₂O =	1	18.02 g H₂O

? mol H₂O = .278 mol H₂O

Now consider the ratios!

.0248 ÷ 1 = .0248 -> so Na₂B₄O₇ is the limiting reactant

.0510 ÷ 1 = .0510 -> so H₂SO₄ is in excess

.278 ÷ 5 = .0556 -> so H₂O is in excess

Determine the mass of excess that remains unreacted. Show your work.

Find the mass of sulfuric acid and the mass of water that will react with the limiting reactant. Then subtract the mass that reacts from 5.00 grams.

? g H₂O =	5.00 g Na₂B₄O₇	1 mol Na₂B₄O₇	5 H₂O	18.02 g H₂O
? g H₂O =	1	201.24 g Na₂B₄O₇	1 Na₂B₄O₇	1 mol H₂O

? g H₂O = 2.24 grams of water react

? g H₂SO₄ =	5.00 g Na₂B₄O₇	1 mol Na₂B₄O₇	1 H₂SO₄	98.08 g H₂SO₄
? g H₂SO₄ =	1	201.24 g Na₂B₄O₇	1 Na₂B₄O₇	1 mol H₂SO₄

? g H₂SO₄ = 2.44 grams of sulfuric acid react

5.00 g – 2.24 g = 2.76 grams of water remain unreacted.

5.00 g – 2.44 g = 2.56 grams of sulfuric acid remain unreacted.

What mass of H₃BO₃ is produced in the reaction? Show your work.

Use the limiting reactant to find the mass of product made by the reaction.

? g H₃BO₃ =	5.00 g Na₂B₄O₇	1 mol Na₂B₄O₇	4 H₃BO₃	61.83g H₃BO₃
? g H₃BO₃ =	1	201.24 g Na₂B₄O₇	1 Na₂B₄O₇	1 mol H₃BO₃

? g H₃BO₃ = 6.14 grams of H₃BO₃ react