## Chemistry 9th Edition

The rate of production of $P_4$ is $6.0\times10^{-4}\space \frac{mol}{L\times s}$. The rate of production of $H_2$ is $3.6\times10^{-3}\space \frac{mol}{L\times s}$.
We are given that $0.0048\space mol$ of $PH_3$ is consumed in a $2.0\space L$ for each second of the reaction $4PH_3\rightarrow P_4+6H_2$. Therefore, the ate at which $PH_3$ is consumed is $0.0024\space \frac{mol}{L\times s}$ because there are $2.0\space L$ of space. Because there is $1\space mol$ of $P_4$ produced for every $4\space mol$ of $PH_3$ consumed, the rate at which $P_4$ is produced is $0.0024\space \frac{mol}{L\times s}\times \frac{1}{4}$ or $6.0\times10^{-4}\space \frac{mol}{L\times s}$. Similarly, because there are $6\space mol$ of $H_2$ produced for every $4\space mol$ of $PH_3$ consumed, the rate at which $P_4$ is produced is $0.0024\space \frac{mol}{L\times s}\times \frac{6}{4}$ or $3.6\times10^{-3}\space \frac{mol}{L\times s}$ Because the amount of liters given has 2 significant figures, both answers also have 2 significant figures.