Stoichiometry Practice Problem Set worksheet featuring questions on iron oxidation to rust and photosynthesis, with conversion factors and spaces for student details.

Let’s solve each part of the stoichiometry problems step by step, using the balanced chemical equations and conversion factors provided.

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## Problem 1: Iron Rusting Reaction

Balanced Equation:
> 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

This tells us:
- 4 moles Fe produce 2 moles Fe₂O₃
- So, mole ratio Fe : Fe₂O₃ = 4 : 2 = 2 : 1

Molar mass of Fe = 55.85 g/mol (standard value — not given in problem, but necessary for calculations)

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a) If 40.0 g of iron react, how many moles of Fe₂O₃ are produced?

Step 1: Convert grams of Fe to moles of Fe
> moles Fe = mass / molar mass = 40.0 g / 55.85 g/mol ≈ 0.7162 mol Fe

Step 2: Use mole ratio to find moles of Fe₂O₃
From equation: 4 mol Fe → 2 mol Fe₂O₃
So, moles Fe₂O₃ = (2/4) × moles Fe = (1/2) × 0.7162 ≈ 0.358 mol Fe₂O₃

✔ Answer: 0.358 moles of Fe₂O₃

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b) If 20.0 g of iron react, how many grams of Fe₂O₃ are produced?

Step 1: Convert grams Fe to moles Fe
> moles Fe = 20.0 g / 55.85 g/mol ≈ 0.3581 mol Fe

Step 2: Use mole ratio to get moles Fe₂O₃
> moles Fe₂O₃ = (2/4) × 0.3581 = 0.17905 mol

Step 3: Convert moles Fe₂O₃ to grams
Molar mass Fe₂O₃ = (2×55.85) + (3×16.00) = 111.7 + 48.0 = 159.7 g/mol

> mass Fe₂O₃ = moles × molar mass = 0.17905 mol × 159.7 g/mol ≈ 28.59 g

✔ Answer: 28.6 g of Fe₂O₃ (rounded to 3 significant figures)

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c) If 10.0 moles of iron react, how many molecules of Fe₂O₃ are produced?

Step 1: Use mole ratio to find moles Fe₂O₃
> moles Fe₂O₃ = (2/4) × 10.0 = 5.00 mol Fe₂O₃

Step 2: Convert moles to molecules using Avogadro’s number
Avogadro’s number = 6.02 × 10²³ molecules/mol

> molecules Fe₂O₃ = 5.00 mol × 6.02 × 10²³ molecules/mol = 3.01 × 10²⁴ molecules

✔ Answer: 3.01 × 10²⁴ molecules of Fe₂O₃

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d) If 3.00 × 10²² atoms of iron react, how many moles of Fe₂O₃ are produced?

Step 1: Convert atoms of Fe to moles of Fe
Use Avogadro’s number: 6.02 × 10²³ atoms/mol

> moles Fe = (3.00 × 10²² atoms) / (6.02 × 10²³ atoms/mol) ≈ 0.04983 mol Fe

Step 2: Use mole ratio to find moles Fe₂O₃
> moles Fe₂O₃ = (2/4) × 0.04983 = 0.02492 mol

✔ Answer: 0.0249 moles of Fe₂O₃ (rounded to 3 sig figs)

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## Problem 2: Photosynthesis Reaction

Balanced Equation:
> 6CO₂(g) + 6H₂O(l) + light energy → C₆H₁₂O₆(aq) + 6O₂(g)

This tells us:
- 6 moles CO₂ → 1 mole glucose (C₆H₁₂O₆)
- 6 moles CO₂ → 6 moles O₂ → so 1:1 ratio CO₂ : O₂

Molar mass glucose = 180.16 g/mol (given)

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a) If 715 g of CO₂ gas is consumed, how many grams of glucose are produced?

Step 1: Molar mass CO₂ = 12.01 + 2×16.00 = 44.01 g/mol

Convert grams CO₂ to moles CO₂:
> moles CO₂ = 715 g / 44.01 g/mol ≈ 16.246 mol CO₂

Step 2: Use mole ratio to find moles glucose
6 mol CO₂ → 1 mol glucose
> moles glucose = 16.246 / 6 ≈ 2.7077 mol

Step 3: Convert moles glucose to grams
> mass glucose = 2.7077 mol × 180.16 g/mol ≈ 487.8 g

✔ Answer: 488 g of glucose (rounded to 3 sig figs)

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b) How many molecules of carbon dioxide are required to produce 5.9 × 10²⁴ molecules of oxygen gas?

From equation: 6 molecules CO₂ → 6 molecules O₂ → so 1:1 ratio

That means: molecules CO₂ needed = molecules O₂ produced

> molecules CO₂ = 5.9 × 10²⁴ molecules

✔ Answer: 5.9 × 10²⁴ molecules of CO₂

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## ✔ Final Answers Summary:

Problem 1:

a) 0.358 mol Fe₂O₃
b) 28.6 g Fe₂O₃
c) 3.01 × 10²⁴ molecules Fe₂O₃
d) 0.0249 mol Fe₂O₃

Problem 2:

a) 488 g glucose
b) 5.9 × 10²⁴ molecules CO₂

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Let me know if you’d like to see the dimensional analysis setup or a diagram for any part!