Stoichiometry Review Answers

Stoichiometry Review Answers

1. a. Na3PO4 Na = 3 mol x 22.99 g/mol = 68.97 g P = 1 mol x 30.97 g/mol = 30.97 g O = 4 mol x 16.00 g/mol = 64.00 g 163.94 g

b. Ca(NO3)2 Ca = 1 mol x 40.08 g/mol = 40.08 g N = 2 mol x 14.01 g/mol = 28.02 g O = 6 mol x 16.00 g/mol = 96.00 g 164.10 g

c. Ca3(PO4)2 Ca = 3 mol x 40.08 g/mol = 120.24 g P = 2 mol x 30.97 g/mol = 61.94 g O = 8 mol x 16.00 g/mol = 128.00 g 310.18 g

d. NaNO3 Na = 1 mol x 22.99 g/mol = 22.99 g N = 1 mol x 14.01 g/mol = 14.01 g O = 3 mol x 16.00 g/mol = 48.00 g 85.00 g

2. The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water.

2 C4H10 + 13O2 8CO2 + 10H2O a. How many moles of water are produced? (Use grams moles)

Molar masses: H2O = 18.02 g/mole O2 = 32.00 g/mole C4H10 = 58.12 g/mole

2.46 g H2O

?

1 mol H2O 18.02 g H2O

= 0.137 mol H2O

b. How many moles of butane are burned? (Use grams moles moles)

2.46 g H2O

?

1 mol H2O 18.02 g H2O

?

2 mol C4H10 10 mol H2O

=

0.0273 mol C4H10

c. Grams of butane burned? (Use grams moles moles grams)

2.46 g H2O

?

1 mol H2O 18.02 g H2O

?

2 mol C4H10 10 mol H2O

?

58.12 g C4H10 1 mol C4H10

=

1.59

g

C4H10

d. Moles of oxygen used?

2.46 g H2O

?

1 mol H2O 18.02 g H2O

?

13 mol O2 10 mol H2O

=

0.177 mol O2

e. Grams of oxygen used?

2.46 g H2O

?

1 mol H2O 18.02 g H2O

?

13 mol O2 10 mol H2O

? 32.00 g O2 1 mol O2

= 5.68 g O2

f. Molecules of water produced?

2.46 g H2O ?

1 mol H2O 18.02 g H2O

?

6.022

?

1023 molecules 1 mol H2O

H2O

=

8.22 ? 1022 molecules H2O

3. Use the following balanced equation. How many grams of sodium sulfate will be produced by the complete reaction of 200.0 grams of sodium hydroxide?

2 NaOH + H2SO4 2 H2O + Na2SO4

200.0 g

x grams

Molar masses: NaOH = 40.00 g/mole Na2SO4 = 142.05 g/mole

moles

moles

200.0 g NaOH

1 mol NaOH ? 40.00 g NaOH

?

1 mol Na2SO4 2 mol NaOH

?

142.05 g Na2SO4 1 mol Na2SO4

=

355.1

g Na2SO4

4. Use the following balanced equation. Calculate the theoretical yield of iron if 16.5 grams of Fe2O3 are completely reacted.

Fe2O3 + 3H2 2Fe + 3H2O

16.5 g

x grams

Molar masses: Fe2O3 = 159.70 Fe = 55.85 g/mole

moles

moles

16.5 g Fe2O3

?

1 mol Fe2O3 159.70 g Fe2O3

2 mol Fe ? 1 mol Fe2O3

55.85 g Fe ? 1 mol Fe = 11.5 g Fe

5. a. Use the following balanced equation. How many grams of lithium nitrate are required to make 250.0 grams of lithium sulfate?

Pb(SO4)2 + 4 LiNO3 Pb(NO3)4 + 2 Li2SO4

x grams

250.0 grams

moles

moles

Molar masses: Li2SO4 = 109.95 g/mole LiNO3 = 68.95 g/mole

250.0 g Li2SO4

?

1 mol Li2SO4 109.95 g Li2SO4

?

4 mol LiNO3 2 mol Li2SO4

?

68.95 g LiNO3 1 mol LiNO3

=

313.6 g LiNO3

b. What is the percent yield if only 195.5 grams of lithium sulfate can be recovered?

195.5 g Percent yield = 250.0 g ?100 = 78.20%

6. a. Use the following balanced equation. Identify the limiting reactant when 1.150 grams of HgO react with 12.46 grams of Cl2.

Convert to moles to get moles available, then calculate moles required:

HgO

1.150 g

1.150 216.6 /

+ 2 Cl2 HgCl2

12.46 g

x g

12.46 70.90 /

+ Cl2O

x g

Molar masses: HgO = 216.6 g/mole Cl2 = 70.90 g/mole HgCl2 = 271.5 g/mole Cl2O = 86.90 g/mole

available 5.309 x 10-3 mol 1.757 x 10-1 mol

required

8.785 x 10-2 mol 1.062 x 10-2 mol

limiting

excess

Calculate moles Cl2 required:

The amount of HgO required to react with all of the Cl2 is more than the amount available. It will run out before all of the Cl2 is used up and, therefore, limits the amount of products made.

5.309 ? 10-3

mol HgO

?

2 mol Cl2 1 mol HgO

=

1.062

x 10-2

mol Cl2

required

Calculate moles HgO required:

1.757 ? 10-1

mol Cl2

?

1 mol HgO 2 mol Cl2

=

8.785 x 10-2

mol HgO required

b. How many grams of each product can be formed?

Use the limiting reactant, HgO, to calculate the grams of each product.

Amount of HgCl2 produced:

1 mol HgO 1.150 g HgO ? 216.6 g HgO

?

1 mol HgCl2 1 mol HgO

?

271.5 g HgCl2 1 mol HgCl2

=

1.441

g

HgCl2

Amount of Cl2O produced:

1.150 g HgO

1 mol HgO ? 216.6 g HgO

?

1 mol Cl2O 1 mol HgO

?

86.90 g Cl2O 1 mol Cl2O

=

0.4614

g

Cl2O

c. How many grams of the excess reactant do not react? Calculate excess moles of Cl2, then convert to grams using molar mass: Moles of Cl2 available ? moles of Cl2 required = 0.1757 moles ? 0.01062 moles = 0.1651 moles Cl2

0.1651

mol

Cl2

?

70.90 g Cl2 1 mol Cl2

=

11.70

g

7. Boron trifluoride reacts with hydrogen gas to produce solid boron and hydrogen fluoride gas. In this reaction 40.0 grams of boron trifluoride are reacted with 5.00 grams of hydrogen gas.

a/b. Write the balanced equation and identify the limiting reactant

2BF3

40.0 g

40.0 67.81 /

+ 3H2 2B + 6HF

5.00 g

x grams

5.00 2.016 /

Molar masses: BF3 = 67.81 g/mole H2 = 2.016 g/mole B = 10.81 g/mol

available 0.590 mol

2.48 mol

required

1.65 mol limiting

0.885 mol excess

Calculate moles BF3 required:

The amount of BF3 required to react with all of the H2 is more than the amount available. It will run out before all of the H2 is used up and, therefore, limits the amount of products made.

2.48 mol H2

?

2 mol BF3 3 mol H2

=

1.65 mol BF3 required

Calculate moles H2 required:

0.590 mol BF3

?

3 mol H2 2 mol BF3

=

0.885 mol H2 required

c. Calculate the mass of boron produced (theoretical yield) using the limiting reactant, BF3.

40.0 g BF3

?

1 mol BF3 67.81 g BF3

2 mol B ? 2 mol BF3

10.81 g B ? 1 mol B = 6.38 g B

d. If the percent yield of boron is 72.6%, what was the actual yield of boron? actual yield

Percent yield = theoretical yield ?100 Rearrange to solve for actual yield:

Actual yield = percent yield ? theoretical yield = 0.726 ? 6.38 g = 4.63 g

8. Examine the unbalanced reaction below:

Na3PO4(aq) + Ca(NO3)2(aq) Ca3(PO4)2(s) + NaNO3(aq)

a. If 30.00 grams of calcium nitrate are reacted with excess sodium phosphate, what is the theoretical yield of

calcium phosphate?

Molar masses:

Balance the equation:

Ca(NO3)2 = 164.10 g/mole

Ca3(PO4)2 = 310.18 g/mole

2Na3PO4(aq) + 3Ca(NO3)2(aq) Ca3(PO4)2(s) + 6NaNO3(aq)

30.00 g

x grams

Solve for mass of the product:

30.00 g Ca(NO3)2

?

1 mol Ca(NO3)2 164.10 g Ca(NO3)2

?

1 mol Ca3(PO4)2 3 mol Ca(NO3)2

?

310.18 g Ca3(PO4)2 1 mol Ca3(PO4)2

=

18.90 g Ca3(PO4)2

b. If 25.45 grams of calcium phosphate are recovered, calculate the percent yield. Note: change the grams of calcium phosphate recovered to 15.45 and then solve

actual yield

15.45 g

Percent yield = theoretical yield ? 100 = 18.90 g ? 100 = 81.75%

9. A chemist combines 50.00 grams of solid magnesium with excess aqueous silver(I) nitrate. Silver precipitates out and magnesium nitrate remains dissolved. The chemist recovers 92.35% of the possible silver produced.

a. Write the balanced equation for the reaction.

Mg + 2AgNO3 2Ag + Mg(NO3)2

50.00 g

x grams

b. What is the actual yield of silver for this reaction? (Hint: what do you have to calculate first?)

Calculate theoretical yield of Ag: 1 mol Mg 2 mol Ag 107.9 g Ag

50.00 g Mg ? 24.31 g Mg ? 1 mol Mg ? 1 mol Ag = 443.9 g Ag

Calculate actual yield:

Actual yield = percent yield ? theoretical yield = 0.9235 ? 443.9 g = 409.9 g

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