A chemist burns 567 liters of propane (C3H8). How many liters of carbon dioxide are made at 1.04 atm and 1,995 degrees Celsius? C3H8 (g) + 5O2 (g) --> 3CO2 (g) + 4H2O (g)

Steps to Solve

1. Write the balanced chemical equation for the combustion of propane.

   The balanced chemical equation for the complete combustion of propane ($C_3H_8$) is: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)$

2. Use the ideal gas law to find the number of moles of propane.

   The ideal gas law is given by $PV = nRT$, where:

   • $P$ = pressure (1.00 atm)
   • $V$ = volume (567 L)
   • $n$ = number of moles
   • $R$ = ideal gas constant (0.0821 L atm / (mol K))
   • $T$ = temperature (298 K)

   Rearrange the ideal gas law to solve for $n$: $n = \frac{PV}{RT}$

   Plug in the values for propane: $n = \frac{(1.00 , \text{atm})(567 , \text{L})}{(0.0821 , \frac{\text{L atm}}{\text{mol K}})(298 , \text{K})} \approx 23.16 , \text{mol}$

3. Determine the number of moles of carbon dioxide produced.

   From the balanced chemical equation, 1 mole of $C_3H_8$ produces 3 moles of $CO_2$. Therefore, the number of moles of $CO_2$ produced is: $n_{CO_2} = 3 \times n_{C_3H_8} = 3 \times 23.16 , \text{mol} \approx 69.48 , \text{mol}$

4. Use the ideal gas law to find the volume of carbon dioxide.

   Using the ideal gas law again, solve for the volume of $CO_2$: $V = \frac{nRT}{P}$

   Plug in the values for carbon dioxide: $V = \frac{(69.48 , \text{mol})(0.0821 , \frac{\text{L atm}}{\text{mol K}})(298 , \text{K})}{(1.00 , \text{atm})} \approx 1699.4 , \text{L}$