Stoichiometry Practice Problem Set featuring questions on iron oxidation and photosynthesis reactions.

Let's solve each problem step by step using stoichiometry principles. We'll use the balanced chemical equations and conversion factors provided.

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Problem 1: Rust Formation

Reaction:
4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

We are given:
- Molar mass of Fe = 55.85 g/mol (standard value, not given but required)
- Molar mass of Fe₂O₃ = (2×55.85) + (3×16.00) = 111.7 + 48.00 = 159.7 g/mol
- Avogadro’s number = 6.02 × 10²³ atoms/mole

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

Step 1: Convert grams of Fe to moles.
$$
\text{Moles of Fe} = \frac{40.0\ \text{g}}{55.85\ \text{g/mol}} = 0.716\ \text{mol Fe}
$$

Step 2: Use mole ratio from balanced equation:
From the reaction:
4 mol Fe → 2 mol Fe₂O₃
So,
$$
\text{Moles of Fe₂O₃} = 0.716\ \text{mol Fe} \times \frac{2\ \text{mol Fe₂O₃}}{4\ \text{mol Fe}} = 0.358\ \text{mol Fe₂O₃}
$$

✔ Answer: 0.358 moles of Fe₂O₃

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

Step 1: Moles of Fe:
$$
\frac{20.0\ \text{g}}{55.85\ \text{g/mol}} = 0.358\ \text{mol Fe}
$$

Step 2: Mole ratio:
$$
0.358\ \text{mol Fe} \times \frac{2\ \text{mol Fe₂O₃}}{4\ \text{mol Fe}} = 0.179\ \text{mol Fe₂O₃}
$$

Step 3: Convert to grams:
$$
0.179\ \text{mol} \times 159.7\ \text{g/mol} = 28.6\ \text{g Fe₂O₃}
$$

✔ Answer: 28.6 grams of Fe₂O₃

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

Step 1: Use mole ratio:
$$
10.0\ \text{mol Fe} \times \frac{2\ \text{mol Fe₂O₃}}{4\ \text{mol Fe}} = 5.00\ \text{mol Fe₂O₃}
$$

Step 2: Convert moles to molecules:
$$
5.00\ \text{mol} \times 6.02 \times 10^{23}\ \text{molecules/mol} = 3.01 \times 10^{24}\ \text{molecules}
$$

✔ 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 to moles of Fe:
$$
\frac{3.00 \times 10^{22}\ \text{atoms}}{6.02 \times 10^{23}\ \text{atoms/mol}} = 0.0498\ \text{mol Fe}
$$

Step 2: Use mole ratio:
$$
0.0498\ \text{mol Fe} \times \frac{2\ \text{mol Fe₂O₃}}{4\ \text{mol Fe}} = 0.0249\ \text{mol Fe₂O₃}
$$

✔ Answer: 0.0249 moles of Fe₂O₃

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Problem 2: Photosynthesis

Reaction:
6CO₂(g) + 6H₂O(l) + light energy → C₆H₁₂O₆(aq) + 6O₂(g)

Given:
- Molar mass of glucose (C₆H₁₂O₆) = 180.16 g/mol
- Molar mass of CO₂ = 44.01 g/mol

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

Step 1: Moles of CO₂:
$$
\frac{715\ \text{g}}{44.01\ \text{g/mol}} = 16.25\ \text{mol CO₂}
$$

Step 2: Use mole ratio:
From the equation:
6 mol CO₂ → 1 mol glucose
$$
16.25\ \text{mol CO₂} \times \frac{1\ \text{mol glucose}}{6\ \text{mol CO₂}} = 2.708\ \text{mol glucose}
$$

Step 3: Convert to grams:
$$
2.708\ \text{mol} \times 180.16\ \text{g/mol} = 487.8\ \text{g glucose}
$$

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

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#### b) How many molecules of CO₂ are required to produce 5.9 × 10²⁴ molecules of O₂?

Step 1: From the balanced equation:
6 mol CO₂ → 6 mol O₂
So, 1:1 mole ratio between CO₂ and O₂.

Therefore, same number of molecules are required.

$$
\text{Molecules of CO₂ needed} = 5.9 \times 10^{24}
$$

✔ 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 a visual explanation or diagram!