Question

How many moles of carbon dioxide are produced when 2.33 liters of propane at 24°C and 67.2 kPa are completely burned in excess oxygen?

Answer 1

To determine how many moles of carbon dioxide are produced when 2.33 liters of propane (C₃H₈) are burned, we first need to follow these steps:

1. Write the balanced chemical equation for the combustion of propane:
\[
C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O
\]
From the equation, we see that 1 mole of propane produces 3 moles of carbon dioxide.

2. Use the ideal gas law to find the number of moles of propane present in the given volume and conditions:
\[
PV = nRT
\]
where:
- \(P\) = pressure (in atm)
- \(V\) = volume (in liters)
- \(n\) = number of moles
- \(R\) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \(T\) = temperature (in Kelvin)

First, convert the pressure from kPa to atm:
\[
P = 67.2 \, \text{kPa} \times \frac{1 \, \text{atm}}{101.325 \, \text{kPa}} \approx 0.663 \, \text{atm}
\]

Convert the temperature from Celsius to Kelvin:
\[
T = 24 + 273.15 = 297.15 \, \text{K}
\]

3. Now use the ideal gas law to find the number of moles \(n\):
\[
n = \frac{PV}{RT} = \frac{(0.663 \, \text{atm}) \times (2.33 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (297.15 \, \text{K})}
\]

Calculating the numerator:
\[
0.663 \times 2.33 \approx 1.544
\]

Calculating the denominator:
\[
0.0821 \times 297.15 \approx 24.475
\]

Now calculate \(n\):
\[
n \approx \frac{1.544}{24.475} \approx 0.063
\]

4. Since 1 mole of propane produces 3 moles of carbon dioxide, the total moles of CO₂ produced is:
\[
\text{moles of } CO_2 = 3 \times n = 3 \times 0.063 \approx 0.189
\]

Thus, approximately **0.189 moles of carbon dioxide** are produced when 2.33 liters of propane are completely burned in excess oxygen.