Question

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

Answer 1

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

Step-by-step 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) * \frac{\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} * V & = & \text{0.3978 mol} * \text{0.083 14 bar}\cdot\text{L}\cdot\text{K}^(-1)\text{mol}^(-1) * \text{273.15 K}\\V & = & \textbf{9.036 L} \\\end{array}\\\text{The volume of phosphorus triiodide is $\large \boxed{\textbf{9.036 L}}$}`