Stoichiometry Practice Problem Set for Chemistry - Free Printable

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:

Reaction:
$$
4\text{Fe}(s) + 3\text{O}_2(g) \rightarrow 2\text{Fe}_2\text{O}_3(s)
$$

We are given:
- Molar mass of Fe = 55.85 g/mol (standard value, not given but needed)
- Molar mass of Fe₂O₃ = (2×55.85) + (3×16.00) = 111.7 + 48.00 = 159.7 g/mol
- Avogadro’s number = $6.02 \times 10^{23}$ atoms/mole

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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.
$$
\text{Moles of Fe} = \frac{40.0\ \text{g}}{55.85\ \text{g/mol}} = 0.716\ \text{mol Fe}
$$

Step 2: Use mole ratio from balanced equation.
From the reaction:
4 mol Fe → 2 mol Fe₂O₃
So, mole ratio:
$$
\frac{2\ \text{mol Fe}_2\text{O}_3}{4\ \text{mol Fe}} = 0.5
$$

$$
\text{Moles of Fe}_2\text{O}_3 = 0.716\ \text{mol Fe} \times \frac{2}{4} = 0.358\ \text{mol Fe}_2\text{O}_3
$$

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

Step 2: Moles of Fe₂O₃:
$$
0.358\ \text{mol Fe} \times \frac{2\ \text{mol Fe}_2\text{O}_3}{4\ \text{mol Fe}} = 0.179\ \text{mol Fe}_2\text{O}_3
$$

Step 3: Convert to grams:
$$
0.179\ \text{mol} \times 159.7\ \text{g/mol} = 28.6\ \text{g Fe}_2\text{O}_3
$$

✔ 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:
$$
10.0\ \text{mol Fe} \times \frac{2\ \text{mol Fe}_2\text{O}_3}{4\ \text{mol Fe}} = 5.00\ \text{mol Fe}_2\text{O}_3
$$

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: $3.01 \times 10^{24}$ molecules of Fe₂O₃

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#### d) If $3.00 \times 10^{22}$ atoms of iron react, how many moles of Fe₂O₃ are produced?

Step 1: Convert atoms of Fe to moles:
$$
\text{Moles of Fe} = \frac{3.00 \times 10^{22}\ \text{atoms}}{6.02 \times 10^{23}\ \text{atoms/mol}} = 0.0498\ \text{mol Fe}
$$

Step 2: Use mole ratio:
$$
0.0498\ \text{mol Fe} \times \frac{2\ \text{mol Fe}_2\text{O}_3}{4\ \text{mol Fe}} = 0.0249\ \text{mol Fe}_2\text{O}_3
$$

✔ Answer: 0.0249 moles of Fe₂O₃

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Problem 2: Photosynthesis Reaction

$$
6\text{CO}_2(g) + 6\text{H}_2\text{O}(l) + \text{light energy} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(aq) + 6\text{O}_2(g)
$$

Given:
- Molar mass of glucose (C₆H₁₂O₆) = 180.16 g/mol
- Molar mass of CO₂ = 44.01 g/mol (not given, but standard)

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

Step 1: Moles of CO₂:
$$
\frac{715\ \text{g}}{44.01\ \text{g/mol}} = 16.25\ \text{mol CO}_2
$$

Step 2: Use mole ratio:
From reaction: 6 mol CO₂ → 1 mol glucose
$$
\text{Moles of glucose} = 16.25\ \text{mol CO}_2 \times \frac{1\ \text{mol glucose}}{6\ \text{mol CO}_2} = 2.708\ \text{mol glucose}
$$

Step 3: Convert to grams:
$$
2.708\ \text{mol} \times 180.16\ \text{g/mol} = 487.7\ \text{g glucose}
$$

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

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#### b) How many molecules of CO₂ are required to produce $5.9 \times 10^{24}$ molecules of O₂?

Step 1: Use mole ratio from reaction:
6 mol CO₂ → 6 mol O₂ → 1:1 ratio

So, same number of moles of CO₂ as O₂ are required.

Step 2: Convert molecules of O₂ to moles:
$$
\text{Moles of O}_2 = \frac{5.9 \times 10^{24}}{6.02 \times 10^{23}} = 9.80\ \text{mol O}_2
$$

Since 1 mol CO₂ : 1 mol O₂, we need 9.80 mol CO₂.

Step 3: Convert to molecules:
$$
9.80\ \text{mol} \times 6.02 \times 10^{23} = 5.89 \times 10^{24}\ \text{molecules CO}_2
$$

✔ Answer: $5.9 \times 10^{24}$ molecules of CO₂ (rounded to 2 sig figs)

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✔ Final Answers Summary:

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#### Problem 1:
a) 0.358 mol Fe₂O₃
b) 28.6 g Fe₂O₃
c) $3.01 \times 10^{24}$ molecules Fe₂O₃
d) 0.0249 mol Fe₂O₃

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#### Problem 2:
a) 488 g glucose
b) $5.9 \times 10^{24}$ molecules CO₂

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