Question

Use the following balanced reaction to solve:

Answer 1

Answer:  60.7 g of $PH_3$ will be formed.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}$    

$\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles$

The balanced chemical reaction is

$P_4(s)+6H_2(g)\rightarrow 4PH_3(g)$

$H_2$ is the limiting reagent as it limits the formation of product and $P_4$ is the excess reagent.

According to stoichiometry :

6 moles of $H_2$ produce = 4 moles of $PH_3$

Thus 2.68 moles of $H_2$ will produce=$\frac{4}{6}\times 2.68=1.79moles$  of $PH_3$

Mass of $PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g$

Thus 60.7 g of $PH_3$ will be formed by reactiong 60 L of hydrogen gas with an excess of $P_4$