When 85.0 g of $$ \mathrm{CH}_{4} $$ are mixed with $$ 160 . \mathrm{g} $$ of $$ \mathrm{O}_{2} $$ what is the maximum amount of $$ \mathrm{CO}_{2} $$ that can be produced? $$ \mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} $$ 7.81 moles 5.00 moles 2.50 moles 5.31 moles

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To determine the maximum amount of $$ \mathrm{CO}_{2} $$ that can be produced, we need to identify the limiting reactant in the reaction:

$$
\mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}
$$

**Step 1: Calculate Moles of Each Reactant**

1. **Moles of $$ \mathrm{CH}_{4} $$:**
   - Molar mass of $$ \mathrm{CH}_{4} $$: $$ 12.01 + 4 \times 1.008 = 16.04 \, \mathrm{g/mol} $$
   - Moles of $$ \mathrm{CH}_{4} $$: $$ \frac{85.0 \, \mathrm{g}}{16.04 \, \mathrm{g/mol}} \approx 5.31 \, \text{mol} $$

2. **Moles of $$ \mathrm{O}_{2} $$:**
   - Molar mass of $$ \mathrm{O}_{2} $$: $$ 32.00 \, \mathrm{g/mol} $$
   - Moles of $$ \mathrm{O}_{2} $$: $$ \frac{160.0 \, \mathrm{g}}{32.00 \, \mathrm{g/mol}} = 5.00 \, \text{mol} $$

**Step 2: Determine the Limiting Reactant**

The balanced equation shows that 1 mole of $$ \mathrm{CH}_{4} $$ requires 2 moles of $$ \mathrm{O}_{2} $$.

- For 5.31 moles of $$ \mathrm{CH}_{4} $$, required $$ \mathrm{O}_{2} $$ = $$ 5.31 \times 2 = 10.62 $$ mol.
- However, only 5.00 mol of $$ \mathrm{O}_{2} $$ is available.

Since $$ \mathrm{O}_{2} $$ is insufficient, it is the **limiting reactant**.

**Step 3: Calculate Moles of $$ \mathrm{CO}_{2} $$ Produced**

From the balanced equation, 2 moles of $$ \mathrm{O}_{2} $$ produce 1 mole of $$ \mathrm{CO}_{2} $$.

- Moles of $$ \mathrm{CO}_{2} $$ produced = $$ \frac{5.00 \, \text{mol} \, \mathrm{O}_{2}}{2} = 2.50 \, \text{mol} $$

**Answer:**  
**2.50 moles** of carbon dioxide can be produced.
2.50 moles of $$ \mathrm{CO}_{2} $$ can be produced.

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