Question

What is the maximum mass of PH3 that can be formed when 62.0g of phosphorus reacts with

Answer 1

Answer: The mass of $PH_3$ produced is 45.22 g

Explanation:

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

• For $P_4$:

Given mass of $P_4$ = 62.0 g

Molar mass of $P_4$ = 124 g/mol

Putting values in equation 1, we get:

$\text{Moles of }P_4=\frac{62.0g}{124g/mol}=0.516mol$

• For $H_2$:

Given mass of $H_2$ = 4.00 g

Molar mass of $H_2$ = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2mol$

The chemical equation follows:

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

By stoichiometry of the reaction:

If 6 moles of hydrogen gas reacts with 1 mole of $P_4$

So, 2 moles of hydrogen gas will react with = $\frac{1}{6}\times 2=0.333mol$ of $P_4$

As the given amount of $P_4$ is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hydrogen gas is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 6 moles of $H_2$ produces 4 mole of $PH_3$

So, 2 moles of $H_2$ will produce = $\frac{4}{6}\times 2=1.33mol$ of $PH_3$

We know, molar mass of $PH_3$ = 34 g/mol

Putting values in equation 1, we get:

$\text{Mass of }PH_3=(1.33mol\times 34g/mol)=45.22g$

Hence, the mass of $PH_3$ produced is 45.22 g