Chemistry - Stoichiometry Worksheet - Edubirdie - Free Printable

Problem 1: Calculate the number of moles of NaOH needed to react with 500.0 g of H₂SO₄.

#### Given:
- Mass of H₂SO₄ = 500.0 g
- Balanced chemical equation:
\[
\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}
\]

#### Step 1: Determine the molar mass of H₂SO₄.
The molar mass of H₂SO₄ is calculated as follows:
\[
\text{Molar mass of H}_2\text{SO}_4 = (2 \times 1.01) + (32.07) + (4 \times 16.00) = 2.02 + 32.07 + 64.00 = 98.09 \, \text{g/mol}
\]

#### Step 2: Calculate the number of moles of H₂SO₄.
Using the formula:
\[
\text{Moles of H}_2\text{SO}_4 = \frac{\text{Mass of H}_2\text{SO}_4}{\text{Molar mass of H}_2\text{SO}_4}
\]
\[
\text{Moles of H}_2\text{SO}_4 = \frac{500.0 \, \text{g}}{98.09 \, \text{g/mol}} \approx 5.10 \, \text{mol}
\]

#### Step 3: Use the stoichiometric ratio to find the moles of NaOH.
From the balanced equation, 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Therefore:
\[
\text{Moles of NaOH} = 2 \times \text{Moles of H}_2\text{SO}_4
\]
\[
\text{Moles of NaOH} = 2 \times 5.10 \, \text{mol} = 10.20 \, \text{mol}
\]

#### Final Answer:
\[
\boxed{10.19 \, \text{mol}}
\]

---

Problem 2: Calculate the mass of NH₃ that can be produced from the reaction of 125 g of NCl₃.

#### Given:
- Mass of NCl₃ = 125 g
- Balanced chemical equation:
\[
\text{NCl}_3 + 3\text{H}_2\text{O} \rightarrow \text{NH}_3 + 3\text{HOCl}
\]

#### Step 1: Determine the molar mass of NCl₃.
The molar mass of NCl₃ is calculated as follows:
\[
\text{Molar mass of NCl}_3 = (1 \times 14.01) + (3 \times 35.45) = 14.01 + 106.35 = 120.36 \, \text{g/mol}
\]

#### Step 2: Calculate the number of moles of NCl₃.
Using the formula:
\[
\text{Moles of NCl}_3 = \frac{\text{Mass of NCl}_3}{\text{Molar mass of NCl}_3}
\]
\[
\text{Moles of NCl}_3 = \frac{125 \, \text{g}}{120.36 \, \text{g/mol}} \approx 1.04 \, \text{mol}
\]

#### Step 3: Use the stoichiometric ratio to find the moles of NH₃.
From the balanced equation, 1 mole of NCl₃ produces 1 mole of NH₃. Therefore:
\[
\text{Moles of NH}_3 = \text{Moles of NCl}_3
\]
\[
\text{Moles of NH}_3 = 1.04 \, \text{mol}
\]

#### Step 4: Calculate the mass of NH₃.
The molar mass of NH₃ is:
\[
\text{Molar mass of NH}_3 = (1 \times 14.01) + (3 \times 1.01) = 14.01 + 3.03 = 17.04 \, \text{g/mol}
\]
Using the formula:
\[
\text{Mass of NH}_3 = \text{Moles of NH}_3 \times \text{Molar mass of NH}_3
\]
\[
\text{Mass of NH}_3 = 1.04 \, \text{mol} \times 17.04 \, \text{g/mol} \approx 17.7 \, \text{g}
\]

#### Final Answer:
\[
\boxed{17.7 \, \text{g}}
\]

---

Problem 3: Identify the limiting reactant and determine the mass of CO₂ that can be produced from the reaction of 25.0 g of C₃H₈ with 75.0 g of O₂.

#### Given:
- Mass of C₃H₈ = 25.0 g
- Mass of O₂ = 75.0 g
- Balanced chemical equation:
\[
\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}
\]

#### Step 1: Determine the molar masses.
- Molar mass of C₃H₈:
\[
\text{Molar mass of C}_3\text{H}_8 = (3 \times 12.01) + (8 \times 1.01) = 36.03 + 8.08 = 44.11 \, \text{g/mol}
\]
- Molar mass of O₂:
\[
\text{Molar mass of O}_2 = 2 \times 16.00 = 32.00 \, \text{g/mol}
\]

#### Step 2: Calculate the moles of C₃H₈ and O₂.
\[
\text{Moles of C}_3\text{H}_8 = \frac{25.0 \, \text{g}}{44.11 \, \text{g/mol}} \approx 0.567 \, \text{mol}
\]
\[
\text{Moles of O}_2 = \frac{75.0 \, \text{g}}{32.00 \, \text{g/mol}} \approx 2.344 \, \text{mol}
\]

#### Step 3: Determine the limiting reactant.
From the balanced equation, 1 mole of C₃H₈ requires 5 moles of O₂. Therefore:
\[
\text{Moles of O}_2 \, \text{required for 0.567 mol of C}_3\text{H}_8 = 0.567 \times 5 = 2.835 \, \text{mol}
\]
Since only 2.344 mol of O₂ is available, O₂ is the limiting reactant.

#### Step 4: Calculate the moles of CO₂ produced using the limiting reactant (O₂).
From the balanced equation, 5 moles of O₂ produce 3 moles of CO₂. Therefore:
\[
\text{Moles of CO}_2 \, \text{produced} = \frac{3}{5} \times \text{Moles of O}_2
\]
\[
\text{Moles of CO}_2 \, \text{produced} = \frac{3}{5} \times 2.344 \approx 1.406 \, \text{mol}
\]

#### Step 5: Calculate the mass of CO₂.
The molar mass of CO₂ is:
\[
\text{Molar mass of CO}_2 = (1 \times 12.01) + (2 \times 16.00) = 12.01 + 32.00 = 44.01 \, \text{g/mol}
\]
Using the formula:
\[
\text{Mass of CO}_2 = \text{Moles of CO}_2 \times \text{Molar mass of CO}_2
\]
\[
\text{Mass of CO}_2 = 1.406 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 61.9 \, \text{g}
\]

#### Final Answer:
\[
\boxed{61.9 \, \text{g}}
\]

---

Problem 4: How many grams of SO₂ are produced when 152 g of CS₂ react with 48.0 g of O₂?

#### Given:
- Mass of CS₂ = 152 g
- Mass of O₂ = 48.0 g
- Balanced chemical equation:
\[
\text{CS}_2 + 3\text{O}_2 \rightarrow \text{CO}_2 + 2\text{SO}_2
\]

#### Step 1: Determine the molar masses.
- Molar mass of CS₂:
\[
\text{Molar mass of CS}_2 = (1 \times 12.01) + (2 \times 32.07) = 12.01 + 64.14 = 76.15 \, \text{g/mol}
\]
- Molar mass of O₂:
\[
\text{Molar mass of O}_2 = 2 \times 16.00 = 32.00 \, \text{g/mol}
\]

#### Step 2: Calculate the moles of CS₂ and O₂.
\[
\text{Moles of CS}_2 = \frac{152 \, \text{g}}{76.15 \, \text{g/mol}} \approx 1.997 \, \text{mol}
\]
\[
\text{Moles of O}_2 = \frac{48.0 \, \text{g}}{32.00 \, \text{g/mol}} = 1.500 \, \text{mol}
\]

#### Step 3: Determine the limiting reactant.
From the balanced equation, 1 mole of CS₂ requires 3 moles of O₂. Therefore:
\[
\text{Moles of O}_2 \, \text{required for 1.997 mol of CS}_2 = 1.997 \times 3 = 5.991 \, \text{mol}
\]
Since only 1.500 mol of O₂ is available, O₂ is the limiting reactant.

#### Step 4: Calculate the moles of SO₂ produced using the limiting reactant (O₂).
From the balanced equation, 3 moles of O₂ produce 2 moles of SO₂. Therefore:
\[
\text{Moles of SO}_2 \, \text{produced} = \frac{2}{3} \times \text{Moles of O}_2
\]
\[
\text{Moles of SO}_2 \, \text{produced} = \frac{2}{3} \times 1.500 = 1.000 \, \text{mol}
\]

#### Step 5: Calculate the mass of SO₂.
The molar mass of SO₂ is:
\[
\text{Molar mass of SO}_2 = (1 \times 32.07) + (2 \times 16.00) = 32.07 + 32.00 = 64.07 \, \text{g/mol}
\]
Using the formula:
\[
\text{Mass of SO}_2 = \text{Moles of SO}_2 \times \text{Molar mass of SO}_2
\]
\[
\text{Mass of SO}_2 = 1.000 \, \text{mol} \times 64.07 \, \text{g/mol} \approx 64.1 \, \text{g}
\]

#### Final Answer:
\[
\boxed{64.1 \, \text{g}}
\]