Question

How many moles of PCl5 can be produced from 19.3 g of P4 (and excess Cl2)?

Answer 1

$$P_4 +10 Cl_2 \rightarrow 4PCl_5$$

$$n=\frac{m}{M}$$

$$n(P_4)=\frac{19.3 g}{123.9\frac{g}{mol}} = 0.156 mol$$

As mole ratio $n(P_4):n(PCl_5)=1:4$, then $n(PCl_5)= 4\times n(P_4)=4\times 0.156 mol = 0.623 mol$