limiting reactant practice answers | Lecture notes Chemistry | Docsity - Free Printable

Let's solve and explain Problem 2 from your chemistry worksheet, since Problem 1 is already solved and explained.

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Problem 2:

Reaction:
$$
\text{CH}_4 + 2\text{H}_2\text{O} \rightarrow 4\text{H}_2(g) + \text{CO}_2(g)
$$

Given:
- 80.0 g of CH₄ (methane)
- 16.3 g of H₂O (water)

Questions:
1. How many liters of hydrogen gas (H₂) can be produced?
2. What is the limiting reactant?

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✔ Step 1: Understand the Concept – Limiting Reactant

The limiting reactant is the one that runs out first and determines how much product can be formed. We need to calculate how much H₂ each reactant can produce and see which one gives less H₂.

We’ll use stoichiometry to convert mass → moles → moles of H₂ → volume of H₂ at STP.

> At STP (Standard Temperature and Pressure), 1 mole of any gas = 22.4 L

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✔ Step 2: Molar Masses

- Molar mass of CH₄ = 12.01 (C) + 4×1.008 (H) = 16.033 g/mol
- Molar mass of H₂O = 2×1.008 + 16.00 = 18.015 g/mol
- Molar mass of H₂ = 2.016 g/mol (but we only need moles and volume)

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✔ Step 3: Calculate Moles of Each Reactant

#### Moles of CH₄:
$$
\frac{80.0\ \text{g}}{16.033\ \text{g/mol}} = 4.990\ \text{moles of CH}_4
$$

#### Moles of H₂O:
$$
\frac{16.3\ \text{g}}{18.015\ \text{g/mol}} = 0.9047\ \text{moles of H}_2\text{O}
$$

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✔ Step 4: Use Stoichiometry to Find H₂ Produced

From the balanced equation:
$$
\text{CH}_4 + 2\text{H}_2\text{O} \rightarrow 4\text{H}_2 + \text{CO}_2
$$

- 1 mol CH₄ → 4 mol H₂
- 2 mol H₂O → 4 mol H₂ → so 1 mol H₂O → 2 mol H₂

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#### From CH₄:
$$
4.990\ \text{mol CH}_4 \times \frac{4\ \text{mol H}_2}{1\ \text{mol CH}_4} = 19.96\ \text{mol H}_2
$$

Convert to volume at STP:
$$
19.96\ \text{mol} \times 22.4\ \text{L/mol} = 447\ \text{L H}_2
$$

✔ So, CH₄ could produce 447 L of H₂

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#### From H₂O:
$$
0.9047\ \text{mol H}_2\text{O} \times \frac{4\ \text{mol H}_2}{2\ \text{mol H}_2\text{O}} = 1.809\ \text{mol H}_2
$$

Convert to volume:
$$
1.809\ \text{mol} \times 22.4\ \text{L/mol} = 40.5\ \text{L H}_2
$$

✔ So, H₂O could produce 40.5 L of H₂

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✔ Step 5: Determine the Limiting Reactant

- CH₄ produces 447 L H₂
- H₂O produces 40.5 L H₂

➡️ Since H₂O produces less H₂, it is the limiting reactant.

> Water (H₂O) will run out first.

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✔ Final Answers:

1. Volume of H₂ produced = 40.5 L
2. Limiting reactant = H₂O (water)
- CH₄ is in excess.

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✔ Alternative Method: Check Required Amounts

You can also determine the limiting reactant by seeing how much of one reactant is needed for the other.

#### How much H₂O is needed for 80.0 g CH₄?
From the reaction: 1 mol CH₄ requires 2 mol H₂O

$$
4.990\ \text{mol CH}_4 \times \frac{2\ \text{mol H}_2\text{O}}{1\ \text{mol CH}_4} = 9.98\ \text{mol H}_2\text{O}
$$

Convert to grams:
$$
9.98\ \text{mol} \times 18.015\ \text{g/mol} = 180.\ \text{g H}_2\text{O required}
$$

But we only have 16.3 g H₂O, which is not enough → H₂O is limiting.

✔️ Confirms our earlier result.

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✔ Summary

| Reactant | Available | Required for CH₄ | Result |
|--------|----------|------------------|--------|
| H₂O | 16.3 g | 180 g | Not enough → Limiting |
| CH₄ | 80.0 g | 80.0 g (excess) | In excess |

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✔ Final Answer Boxed:

> Answer:
> - The limiting reactant is water (H₂O).
> - 40.5 liters of hydrogen gas can be produced.

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Let me know if you'd like a visual diagram or explanation of why this matters in real-world reactions!