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(D) Assume $100 	ext{ moles}$ of the mixture. Then,$n_A = 50 	ext{ mol}$,$n_B = 30 	ext{ mol}$,and $n_C = 20 	ext{ mol}$.In reaction $1$: $A + 2B

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$P_1$ will be completely exhausted.

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(D) Assume $100 	ext{ moles}$ of the mixture. Then,$n_A = 50 	ext{ mol}$,$n_B = 30 	ext{ mol}$,and $n_C = 20 	ext{ mol}$.In reaction $1$: $A + 2B ightarrow P_1$.For $50 	ext{ mol}$ of $A$,we need $100 	ext{ mol}$ of $B$. Since we only have $30 	ext{ mol}$ of $B$,$B$ is the limiting reagent.$30 	ext{ mol}$ of $B$ will react with $15 	ext{ mol}$ of $A$ to produce $15 	ext{ mol}$ of $P_1$.Remaining $A = 50 - 15 = 35 	ext{ mol}$.In reaction $2$: $4P_1 + 3C ightarrow P_2$.We have $15 	ext{ mol}$ of $P_1$ and $20 	ext{ mol}$ of $C$.For $15 	ext{ mol}$ of $P_1$,the required amount of $C$ is $\frac{3}{4} 	imes 15 = 11.25 	ext{ mol}$.Since we have $20 	ext{ mol}$ of $C$,$P_1$ is the limiting reagent in the second reaction.$15 	ext{ mol}$ of $P_1$ will react with $11.25 	ext{ mol}$ of $C$ to produce $P_2$.Remaining $C = 20 - 11.25 = 8.75 	ext{ mol}$.Remaining $A = 35 	ext{ mol}$.Since $P_1$ is completely consumed in the second reaction,$P_1$ will be completely exhausted.

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