Question

How many grams of $P{F}_5$ can be formed from 9.46 g of $P{F}_3$ and 9.42 g of $Xe{F}_4$ in the following reaction? (Molar mass of Xe =131 g/mol)
$2P{F}_3+Xe{F}_4\to 2P{F}_5+Xe$

• A. 11.46 g
• B. 20.63 g
• C. 5.19 g
• D. 14.87 g

Answer 1

The correct option is A 11.46 g

$2P{F}_3+Xe{F}_4\to 2P{F}_5+Xe$
Moles of $P{F}_3=\frac{9.46}{88}=0.1075$ moles
Moles of $Xe{F}_4=\frac{9.42}{207}=0.0455$ moles
2 moles of $P{F}_3$ react with 1 mole of $Xe{F}_4$
0.1075 moles will react with $\frac{1}{2}\times 0.1075=0.0538$ moles of $Xe{F}_4$
Therefore, $Xe{F}_4$ is the limiting reagent.
1 mole of $Xe{F}_4$ produces 2 moles of $P{F}_5$
0.0455 moles will produce $2\times 0.0455=0.091$ moles of $P{F}_5$
Mass of $P{F}_5=0.091\times \text{molar mass}$ of $P{F}_5=(0.091\times 126)$ g
Mass of $P{F}_5$ = 11.46 g