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Stoichiometry Calculation Practice Worksheet featuring four chemistry problems with balanced equations and answers.

Stoichiometry Calculation Practice Worksheet with four problems involving chemical reactions and calculations, including moles, mass, limiting reactants, and product yields.

Stoichiometry Calculation Practice Worksheet with four problems involving chemical reactions and calculations, including moles, mass, limiting reactants, and product yields.

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Show Answer Key & Explanations Step-by-step solution for: Chemistry - Stoichiometry Worksheet - Edubirdie
1. Calculate the number of moles of NaOH needed to react with 500.0 g of H₂SO₄.
- Molar mass of H₂SO₄ = 98.08 g/mol.
- Moles of H₂SO₄ = 500.0 g / 98.08 g/mol ≈ 5.098 mol.
- From the balanced equation, 1 mol H₂SO₄ reacts with 2 mol NaOH.
- Moles of NaOH = 2 × 5.098 mol ≈ 10.196 mol.
- Rounded to two decimal places: 10.19 mol.

2. Calculate the mass of NH₃ produced from 125 g of NCl₃.
- Molar mass of NCl₃ = 120.36 g/mol.
- Moles of NCl₃ = 125 g / 120.36 g/mol ≈ 1.0385 mol.
- From the balanced equation, 1 mol NCl₃ produces 1 mol NH₃.
- Moles of NH₃ = 1.0385 mol.
- Molar mass of NH₃ = 17.03 g/mol.
- Mass of NH₃ = 1.0385 mol × 17.03 g/mol ≈ 17.69 g.
- Rounded to one decimal place: 17.7 g.

3. Identify the limiting reactant and determine the mass of CO₂ produced from 25.0 g C₃H₈ and 75.0 g O₂.
- Molar mass of C₃H₈ = 44.10 g/mol; moles = 25.0 g / 44.10 g/mol ≈ 0.5669 mol.
- Molar mass of O₂ = 32.00 g/mol; moles = 75.0 g / 32.00 g/mol ≈ 2.3438 mol.
- From the balanced equation, 1 mol C₃H₈ requires 5 mol O₂.
- O₂ required for 0.5669 mol C₃H₈ = 5 × 0.5669 ≈ 2.8345 mol (but only 2.3438 mol available).
- Thus, O₂ is limiting.
- From the equation, 5 mol O₂ produce 3 mol CO₂.
- Moles of CO₂ = (3/5) × 2.3438 mol ≈ 1.4063 mol.
- Molar mass of CO₂ = 44.01 g/mol.
- Mass of CO₂ = 1.4063 mol × 44.01 g/mol ≈ 61.89 g.
- Rounded to one decimal place: 61.9 g.

4. Calculate grams of SO₂ produced when 152 g CS₂ react with 48.0 g O₂.
- Molar mass of CS₂ = 76.14 g/mol; moles = 152 g / 76.14 g/mol ≈ 1.996 mol.
- Molar mass of O₂ = 32.00 g/mol; moles = 48.0 g / 32.00 g/mol = 1.500 mol.
- From the balanced equation, 1 mol CS₂ requires 3 mol O₂.
- O₂ required for 1.996 mol CS₂ = 3 × 1.996 ≈ 5.988 mol (but only 1.500 mol available).
- Thus, O₂ is limiting.
- From the equation, 3 mol O₂ produce 2 mol SO₂.
- Moles of SO₂ = (2/3) × 1.500 mol = 1.000 mol.
- Molar mass of SO₂ = 64.07 g/mol.
- Mass of SO₂ = 1.000 mol × 64.07 g/mol = 64.07 g.
- Rounded to one decimal place: 64.1 g.
Parent Tip: Review the logic above to help your child master the concept of basic stoichiometry worksheet.
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