To determine the volume of $\text{CO}_2$ produced, we need to follow these steps:

1. Write the balanced chemical equation for the combustion of methane:

   $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

   This equation tells us that 1 mole of methane ($\text{CH}_4$) produces 1 mole of carbon dioxide ($\text{CO}_2$).

2. Calculate the moles of methane ($\text{CH}_4$) given the mass:

   The molar mass of methane ($\text{CH}_4$) is: $12.01 \, \text{g/mol} + (4 \times 1.008 \, \text{g/mol}) = 16.04 \, \text{g/mol}$

   Convert 7.50 kg of methane to grams and then to moles: $7.50 \, \text{kg} = 7500 \, \text{g}$ $\text{Moles of } \text{CH}_4 = \frac{7500 \, \text{g}}{16.04 \, \text{g/mol}} \approx 467.67 \, \text{mol}$

3. Determine the moles of $\text{CO}_2$ produced:

   From the balanced equation, 1 mole of $\text{CH}_4$ produces 1 mole of $\text{CO}_2$. Therefore: $\text{Moles of } \text{CO}_2 = 467.67 \, \text{mol}$

4. Use the Ideal Gas Law to find the volume of $\text{CO}_2$ at the specified conditions:

   The Ideal Gas Law is: $PV = nRT$ where:

   • $P = 0.956 \, \text{atm}$
   • $n = 467.67 \, \text{mol}$
   • $R = 0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K})$
   • $T = 20.0^\circ \text{C} = 293.15 \, \text{K}$

   Rearrange to solve for $V$: $V = \frac{nRT}{P}$

   Substitute the values: $V = \frac{(467.67 \, \text{mol}) \times (0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}) \times 293.15 \, \text{K}}{0.956 \, \text{atm}}$

   Calculating this gives: $V \approx 11,323.6 \, \text{L}$

Answer: The volume of carbon dioxide produced is approximately 11,323.6 liters.

Would you like further details or clarification on any part of this solution?

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Related Questions:

1. How would the volume change if the temperature were increased to 50°C?
2. How would this calculation differ at a pressure of 1.5 atm?
3. What is the volume of water vapor produced under the same conditions?
4. How does the Ideal Gas Law account for non-ideal gas behavior?
5. How would the volume change if we started with 10 kg of methane?

Tip: Always check that the temperature in gas law calculations is converted to Kelvin.