Question

How many moles of carbon dioxide are produced when 4.00 L of propane are reacted at STP?

C3H8 + 5O2 —> 3CO2 + 4H2O

Answer 1

`\large \boxed{\text{0.528 mol}}`

Step-by-step 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:

`\begin{array}{rcl}\\pV &=& nRT\\\text{1 bar}*\text{4.00 L} & = & n * 0.08314 \text{ bar}\cdot\text{L}\cdot\text{K}^(-1)\text{mol}^(-1) * \text{ 273.15 K}\\4.00 & = & 22.71n \text{ mol}^(-1)\\n & = & \frac{4.00}{22.71 \text{ mol}^(-1)}\\\\& = & 0.1761 \text{ mol}\\\end{array}`

(b) Moles of CO₂

The molar ratio is 3 mol CO₂/1 mol C₃H₈.

`\text{Moles of CO}_(2) = \text{0.1761 mol C$_(3)$H}_(8) * \frac{\text{3 mol CO}_(2)}{\text{1 mol C$_(3)$H}_(8)} = \textbf{0.528 mol CO}_(2)\\\\\text{The reaction produces $\large \boxed{\textbf{0.528 mol CO}_(2)}$}`