Gas Stoichiometry Practice Worksheet Answer Key

You’ve uploaded an image of a Gas Stoichiometry Practice 1 Answer Key, which contains 5 stoichiometry problems involving gases at STP (Standard Temperature and Pressure: 0°C and 1 atm). Below, I’ll solve each problem step-by-step with clear explanations — even though the image already shows the answers, I’ll walk through the reasoning so you understand *how* to get there.

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## 🔹 Problem 1:
> Calculate the volume of hydrogen gas that will react with exactly 4.61 L of oxygen gas using the equation:
> 2 H₂(g) + O₂(g) → 2 H₂O(g)
> Assume there is no change in temperature or pressure.

✔ Solution:

This is a volume-to-volume gas stoichiometry problem at constant T and P. At STP, we can use the mole ratio from the balanced equation as a volume ratio (Avogadro’s Law: equal volumes of gases at same T & P contain equal moles).

From the balanced equation:
2 volumes H₂ : 1 volume O₂

So, if 1 L of O₂ reacts with 2 L of H₂, then:

> Volume H₂ = 4.61 L O₂ × (2 L H₂ / 1 L O₂) = 9.22 L H₂

✔ Answer: 9.22 L H₂

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## 🔹 Problem 2:
> What volume of CO₂ gas will be produced from the reaction of 2 C₃H₈(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(g)?

Wait — this question seems incomplete. It doesn’t specify how much propane (C₃H₈) or oxygen is given. Looking at the answer key, it says:

> 17.5 L CO₂ (from 8.75 L C₃H₈)

So let’s assume the problem meant:
> “What volume of CO₂ gas will be produced from 8.75 L of C₃H₈?”

✔ Solution:

Balanced equation:
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)
*(Note: The equation written in the image has coefficients doubled — 2 C₃H₈ + 10 O₂ → 6 CO₂ + 8 H₂O — but the ratio is the same.)*

Volume ratio:
1 volume C₃H₈ → 3 volumes CO₂

So:

> Volume CO₂ = 8.75 L C₃H₈ × (3 L CO₂ / 1 L C₃H₈) = 26.25 L CO₂

Wait — the answer key says 17.5 L CO₂. That suggests maybe they used a different ratio or miswrote the problem.

Let me check the image again — it says:

> 17.5 L CO₂ ← from 8.75 L C₃H₈

But according to the balanced equation (even with doubled coefficients), 2 mol C₃H₈ → 6 mol CO₂ → ratio 1:3.

So 8.75 L × 3 = 26.25 L.

Hmm — perhaps the problem was actually:
> “What volume of CO₂ is produced from 5.83 L O₂?” or something else?

Alternatively, maybe the image has a typo. Let’s trust the math: if 8.75 L C₃H₈ → 17.5 L CO₂, that implies a 1:2 ratio — which is wrong for propane combustion.

✔ Correct Answer should be: 26.25 L CO₂ (if starting with 8.75 L C₃H₈)

But since the answer key says 17.5 L, perhaps the original problem was different. For now, I’ll note the discrepancy and proceed.

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## 🔹 Problem 3:
> What volume of hydrogen gas at STP will be produced from the reaction of 36.5 g of zinc with hydrochloric acid?
> Zn(s) + 2 HCl(aq) → H₂(g) + ZnCl₂(aq)

✔ Solution:

This is a mass → volume stoichiometry problem.

Step 1: Convert mass of Zn to moles.

Molar mass Zn = 65.38 g/mol (≈65.4 g/mol)

> Moles Zn = 36.5 g / 65.4 g/mol ≈ 0.558 mol Zn

Step 2: Use mole ratio from balanced equation.

1 mol Zn → 1 mol H₂

So, moles H₂ = 0.558 mol

Step 3: Convert moles of H₂ to volume at STP.

At STP, 1 mol gas = 22.4 L

> Volume H₂ = 0.558 mol × 22.4 L/mol ≈ 12.5 L H₂

✔ Answer: 12.5 L H₂

*(Matches answer key)*

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## 🔹 Problem 4:
> How many grams of NaCl are produced from the reaction of 12.6 L of chlorine gas at STP with excess sodium?
> Cl₂(g) + 2 Na(s) → 2 NaCl(s)

✔ Solution:

This is a volume → mass stoichiometry problem.

Step 1: Convert volume of Cl₂ at STP to moles.

> Moles Cl₂ = 12.6 L / 22.4 L/mol ≈ 0.5625 mol Cl₂

Step 2: Use mole ratio.

1 mol Cl₂ → 2 mol NaCl

> Moles NaCl = 0.5625 × 2 = 1.125 mol NaCl

Step 3: Convert moles NaCl to grams.

Molar mass NaCl = 58.44 g/mol

> Mass NaCl = 1.125 mol × 58.44 g/mol ≈ 65.7 g NaCl

✔ Answer: 65.7 g NaCl

*(Matches answer key)*

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## 🔹 Problem 5:
> In order to produce 50.0 liters of ammonia gas (NH₃) at STP, how many grams of both hydrogen and nitrogen would you need?
> N₂(g) + 3 H₂(g) → 2 NH₃(g)

✔ Solution:

We need to find mass of N₂ and H₂ required to make 50.0 L NH₃ at STP.

Step 1: Convert volume NH₃ to moles.

> Moles NH₃ = 50.0 L / 22.4 L/mol ≈ 2.232 mol NH₃

Step 2: Use mole ratios to find moles of reactants.

From balanced equation:
1 mol N₂ → 2 mol NH₃
→ Moles N₂ needed = 2.232 mol NH₃ × (1 mol N₂ / 2 mol NH₃) = 1.116 mol N₂

3 mol H₂ → 2 mol NH₃
→ Moles H₂ needed = 2.232 mol NH₃ × (3 mol H₂ / 2 mol NH₃) = 3.348 mol H₂

Step 3: Convert moles to grams.

- Molar mass N₂ = 28.02 g/mol
→ Mass N₂ = 1.116 mol × 28.02 g/mol ≈ 31.3 g N₂

- Molar mass H₂ = 2.016 g/mol
→ Mass H₂ = 3.348 mol × 2.016 g/mol ≈ 6.75 g H₂

✔ Answers: 31.3 g N₂ and 6.75 g H₂

*(Matches answer key)*

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## 🧠 Summary of Key Concepts:

- At STP, 1 mole of any gas = 22.4 L
- For gas-gas reactions at same T & P, you can use volume ratios = mole ratios (Avogadro’s Law)
- Always start with a balanced chemical equation
- Use dimensional analysis (unit conversion) to go from:
- Volume → moles → moles → mass (or vice versa)
- Pay attention to limiting reactants (here, usually one reactant is in excess)

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## ❗ Note on Problem 2:
There appears to be a mismatch between the problem statement and the answer provided. If the problem intended to say “from 8.75 L C₃H₈”, then the correct answer is 26.25 L CO₂, not 17.5 L. Double-check the original problem wording.

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✔ All other problems are solved correctly per the answer key and standard stoichiometry rules.

Let me know if you’d like to see any step expanded or want practice problems!