Answer:
[tex]\large \boxed{\text{0.528 mol}}[/tex]
Explanation:
We will need a balanced chemical equation with moles and volumes, so, let's gather all the information in one place.
C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O
V/L: 4.00
(a) Moles of C₃H₈
We can use the Ideal Gas Law:
pV = nRT
Data:
p = 1 bar
V = 4.00 L
T = 273.15 K
Calculation:
[tex]\begin{array}{rcl}\\pV &=& nRT\\\text{1 bar}\times\text{4.00 L} & = & n \times 0.08314 \text{ bar}\cdot\text{L}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{ 273.15 K}\\4.00 & = & 22.71n \text{ mol}^{-1}\\n & = & \dfrac{4.00}{22.71 \text{ mol}^{-1}}\\\\& = & 0.1761 \text{ mol}\\\end{array}[/tex]
(b) Moles of CO₂
The molar ratio is 3 mol CO₂/1 mol C₃H₈.
[tex]\text{Moles of CO}_{2} = \text{0.1761 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} = \textbf{0.528 mol CO}_{2}\\\\\text{The reaction produces $\large \boxed{\textbf{0.528 mol CO}_{2}}$}[/tex]
