Question

You have 163.8 g of phosphorus triiodide gas. What volume of gas do you have at STP?

Answer 1

Answer:

$\large \boxed{\text{9.036 L}}$

Explanation:

We can convert the mass to moles and then use the Ideal Gas Law:

pV = nRT

1. Moles of PI₃:

$n = \text{163.8 g PI}_{3} \times \dfrac{\text{1 mol PI}_{3}}{\text{411.69 g PI}_{3}} = \text{0.3978 mol PI}_{3}$

2. Volume of gas

STP = 273.15 K and 1 bar

$\begin{array}{rcl}\text{1 bar} \times V & = & \text{0.3978 mol} \times \text{0.083 14 bar}\cdot\text{L}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{273.15 K}\\V & = & \textbf{9.036 L} \\\end{array}\\\text{The volume of phosphorus triiodide is \large \boxed{\textbf{9.036 L}} }$