Question

The mass of $P_4{O}_{10}$ produced, if $440$ g of $P_4{S}_3$ is mixed with $384gm$ of $O_2$, is:

$P_4{S}_3+O_2\to P_4{O}_{10}+S{O}_2$

• A. $568gm$
• B. $426gm$
• C. $284gm$
• D. $396gm$

Answer 1

Answer: (A)

$568gm$

The balanced chemical equation is $P_4{S}_3+8O_2\to P_4{O}_{10}+3S{O}_2$.

The molar masses of $P_4{S}_3,O_2$ and $P_4{O}_{10}$ are $220$ g/ mol, $32$ g/mol and $283.9$ g/mol, respectively.

The number of moles of $P_4{S}_3$ and $O_2$ are $\frac{440}{220}=2$ and $\frac{384}{32}=12$, respectively.

1 mol of $P_4{S}_3$ reacts with 8 moles of $O_2$. So, 2 mol of $P_4{S}_3$ reacts with 16 moles of $O_2$.

Thus, $P_4{S}_3$ is the limiting reagent since there is only 12 moles of $O_2$.

$1$ moles of $P_4{S}_3$ will react with $8$ moles of oxygen to form $1$ moles of $P_4{O}_{10}$.

The mass of $P_4{O}_{10}$ produced is $\frac{12\times 284}{8}=426$ g.

Hence, the correct option is $B$.