Question

What mass of $P_4{O}_{10}$ will be produced by the combustion of 2.0 g fo $P_4$ with 2.0 g of $O_2$?

• A. 1.04 g
• B. 0.52 g
• C. 2.04 g
• D. 3.55 g

Answer 1

The correct option is D 3.55 g

The molar masses of $P_4$, $O_2$ and $P_4{O}_{10}$ are $4\times 31=124$ g/mol, $2\times 16=32$ g/mol and $4\left(31\right)+10\left(16\right)=284$ g/mol respectively.
2.0 g of $P_4$ corresponds to $\frac{2.0}{124}=0.016$ moles.
2.0 g of $O_2$ corresponds to $\frac{2.0}{32}=0.0625$ moles.
1 mole of $P_4$ combines with 5 moles of $P_4{O}_{10}$
Hence, 0.016 moles of $P_4$ will combine with $5\times 0.016=0.080$ moles of $O_2$
However, only 0.0625 moles of $O_2$ are present.
Thus, $O_2$ is the limiting reagent.
5 moles of $O_2$ will produce 1 mole of $P_4{O}_{10}$.
Hence, 0.0625 moles of $O_2$ will produce $\frac{0.0625}{5}=0.0125$ moles of $P_4{O}_{10}$.
This corresponds to $0.0125\times 284=3.55$ g of $P_4{O}_{10}$.