Question

What weight of each compound will be obtained when $0.450$ g of $P_4{O}_{10}$ and 1.50 g of $PC{l}_5$ are allowed to react completely?
Given reaction: $6PC{l}_5+P_4{O}_{10}\to 10POC{l}_3$

• A. $POC{l}_3=1.84$ g; $P_4{O}_{10}=0.11$ g
• B. $POC{l}_3=5.00$ g; $P_4{O}_{10}=0.24$ g
• C. $POC{l}_3=2.68$ g; $P_4{O}_{10}=0.36$ g
• D. $POC{l}_3=0.79$ g; $P_4{O}_{10}=0.09$ g

Answer 1

The correct option is A $POC{l}_3=1.84$ g; $P_4{O}_{10}=0.11$ g

$P_4{O}_{10}+6PC{l}_5\to 10POC{l}_3$
6 moles of $PC{l}_5$ react with 1 mole of $P_4{O}_{10}$
$(6\times 208.5)$ g of $PC{l}_5$ react with 284 g of $P_4{O}_{10}$
1.5 g of $PC{l}_5$ will react with $\frac{284}{6\times 208.5}\times 1.5$ = 0.340 g of $P_4{O}_{10}$
Therefore, $PC{l}_5$ is the limiting reagent.
6 moles of $PC{l}_5$ produce 10 moles of $POC{l}_3$
$(6\times 208.5)$ g of $PC{l}_5$ will produce $(10\times 153.5)$g of $POC{l}_3$
So, 1.5 g will produce $=\frac{1535}{6\times 208.5}\times 1.5=1.84$ g of $POC{l}_3$
Mass of excess $P_4{O}_{10}=0.450-0.340=0.11$ g