Chemistry worksheet on stoichiometry mass-mass calculations.

Here are the step-by-step solutions for each problem on the worksheet. I have used dimensional analysis (the factor-label method) as requested, checking significant figures and units along the way.

Problem 1

Question: What mass of iron is needed to react with sulfur in order to produce 96 grams of iron (III) sulfide?
Equation: $2 \text{Fe} + 3 \text{S} \rightarrow \text{Fe}_2\text{S}_3$

1. Molar Masses:
* Iron (Fe): $55.85 \text{ g/mol}$
* Iron (III) Sulfide ($\text{Fe}_2\text{S}_3$): $(2 \times 55.85) + (3 \times 32.06) = 111.7 + 96.18 = 207.88 \text{ g/mol}$

2. Dimensional Analysis Setup:
$$ 96 \text{ g } \text{Fe}_2\text{S}_3 \times \frac{1 \text{ mol } \text{Fe}_2\text{S}_3}{207.88 \text{ g } \text{Fe}_2\text{S}_3} \times \frac{2 \text{ mol Fe}}{1 \text{ mol } \text{Fe}_2\text{S}_3} \times \frac{55.85 \text{ g Fe}}{1 \text{ mol Fe}} $$

3. Calculation:
$$ \frac{96 \times 1 \times 2 \times 55.85}{207.88 \times 1 \times 1} = \frac{10723.2}{207.88} \approx 51.58 \text{ g} $$

4. Significant Figures: The given value "96" has 2 significant figures. We round the answer to 2 significant figures.
* Result: $52 \text{ g}$

***

Problem 2

Question: What mass of zinc is needed to react with 23.1 g of phosphoric acid?
Equation: $3 \text{Zn} + 2 \text{H}_3(\text{PO}_4) \rightarrow 3 \text{H}_2 + \text{Zn}_3(\text{PO}_4)_2$

1. Molar Masses:
* Zinc (Zn): $65.38 \text{ g/mol}$
* Phosphoric Acid ($\text{H}_3\text{PO}_4$): $(3 \times 1.008) + 30.97 + (4 \times 16.00) = 3.024 + 30.97 + 64.00 = 97.99 \text{ g/mol}$

2. Dimensional Analysis Setup:
$$ 23.1 \text{ g } \text{H}_3\text{PO}_4 \times \frac{1 \text{ mol } \text{H}_3\text{PO}_4}{97.99 \text{ g } \text{H}_3\text{PO}_4} \times \frac{3 \text{ mol Zn}}{2 \text{ mol } \text{H}_3\text{PO}_4} \times \frac{65.38 \text{ g Zn}}{1 \text{ mol Zn}} $$

3. Calculation:
$$ \frac{23.1 \times 1 \times 3 \times 65.38}{97.99 \times 2 \times 1} = \frac{4537.37}{195.98} \approx 23.15 \text{ g} $$

4. Significant Figures: The given value "23.1" has 3 significant figures.
* Result: $23.2 \text{ g}$

***

Problem 3

Question: If 48 g of magnesium reacts with oxygen gas, how many grams of magnesium oxide will be formed?
Equation: $2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO}$

1. Molar Masses:
* Magnesium (Mg): $24.31 \text{ g/mol}$
* Magnesium Oxide (MgO): $24.31 + 16.00 = 40.31 \text{ g/mol}$

2. Dimensional Analysis Setup:
$$ 48 \text{ g Mg} \times \frac{1 \text{ mol Mg}}{24.31 \text{ g Mg}} \times \frac{2 \text{ mol MgO}}{2 \text{ mol Mg}} \times \frac{40.31 \text{ g MgO}}{1 \text{ mol MgO}} $$

3. Calculation:
$$ \frac{48 \times 1 \times 2 \times 40.31}{24.31 \times 2 \times 1} = \frac{3869.76}{48.62} \approx 79.59 \text{ g} $$

4. Significant Figures: The given value "48" has 2 significant figures.
* Result: $80. \text{ g}$ (or $8.0 \times 10^1 \text{ g}$)

***

Problem 4

Question: How many grams of copper (I) chloride can be produced from the reaction of 73.5 g of copper (I) oxide with hydrochloric acid?
Equation: $\text{Cu}_2\text{O} + 2 \text{HCl} \rightarrow 2 \text{CuCl} + \text{H}_2\text{O}$

1. Molar Masses:
* Copper (I) Oxide ($\text{Cu}_2\text{O}$): $(2 \times 63.55) + 16.00 = 143.10 \text{ g/mol}$
* Copper (I) Chloride (CuCl): $63.55 + 35.45 = 99.00 \text{ g/mol}$

2. Dimensional Analysis Setup:
$$ 73.5 \text{ g } \text{Cu}_2\text{O} \times \frac{1 \text{ mol } \text{Cu}_2\text{O}}{143.10 \text{ g } \text{Cu}_2\text{O}} \times \frac{2 \text{ mol CuCl}}{1 \text{ mol } \text{Cu}_2\text{O}} \times \frac{99.00 \text{ g CuCl}}{1 \text{ mol CuCl}} $$

3. Calculation:
$$ \frac{73.5 \times 1 \times 2 \times 99.00}{143.10 \times 1 \times 1} = \frac{14553}{143.10} \approx 101.69 \text{ g} $$

4. Significant Figures: The given value "73.5" has 3 significant figures.
* Result: $102 \text{ g}$

***

Problem 5

Question: If 11.5 grams of chlorine reacts with aluminum, how many grams of aluminum chloride will be formed?
Equation: $2 \text{Al} + 3 \text{Cl}_2 \rightarrow 2 \text{AlCl}_3$

1. Molar Masses:
* Chlorine Gas ($\text{Cl}_2$): $2 \times 35.45 = 70.90 \text{ g/mol}$
* Aluminum Chloride ($\text{AlCl}_3$): $26.98 + (3 \times 35.45) = 26.98 + 106.35 = 133.33 \text{ g/mol}$

2. Dimensional Analysis Setup:
$$ 11.5 \text{ g } \text{Cl}_2 \times \frac{1 \text{ mol } \text{Cl}_2}{70.90 \text{ g } \text{Cl}_2} \times \frac{2 \text{ mol AlCl}_3}{3 \text{ mol } \text{Cl}_2} \times \frac{133.33 \text{ g AlCl}_3}{1 \text{ mol AlCl}_3} $$

3. Calculation:
$$ \frac{11.5 \times 1 \times 2 \times 133.33}{70.90 \times 3 \times 1} = \frac{3066.59}{212.7} \approx 14.417 \text{ g} $$

4. Significant Figures: The given value "11.5" has 3 significant figures.
* Result: $14.4 \text{ g}$

***

Problem 6

Question: How many grams of silver bromide are produced when 206 grams of cobalt (III) nitrate are produced?
Equation: $\text{CoBr}_3 + 3 \text{Ag}(\text{NO}_3) \rightarrow 3 \text{AgBr} + \text{Co}(\text{NO}_3)_3$

1. Molar Masses:
* Cobalt (III) Nitrate ($\text{Co}(\text{NO}_3)_3$): $58.93 + 3(14.01 + 3(16.00)) = 58.93 + 3(62.01) = 58.93 + 186.03 = 244.96 \text{ g/mol}$
* Silver Bromide (AgBr): $107.87 + 79.90 = 187.77 \text{ g/mol}$

2. Dimensional Analysis Setup:
$$ 206 \text{ g } \text{Co}(\text{NO}_3)_3 \times \frac{1 \text{ mol } \text{Co}(\text{NO}_3)_3}{244.96 \text{ g } \text{Co}(\text{NO}_3)_3} \times \frac{3 \text{ mol AgBr}}{1 \text{ mol } \text{Co}(\text{NO}_3)_3} \times \frac{187.77 \text{ g AgBr}}{1 \text{ mol AgBr}} $$

3. Calculation:
$$ \frac{206 \times 1 \times 3 \times 187.77}{244.96 \times 1 \times 1} = \frac{116041.86}{244.96} \approx 473.71 \text{ g} $$

4. Significant Figures: The given value "206" has 3 significant figures.
* Result: $474 \text{ g}$

──────────────────────────────────────

Final Answer:
1. 52 g Fe
2. 23.2 g Zn
3. 80. g MgO
4. 102 g CuCl
5. 14.4 g AlCl₃
6. 474 g AgBr