Question

12.Use the equation below to determine the maximum number of grams of PH3 that can be formed when 8.2 g of phosphorus reacts with 4.0 g of hydrogen to form PH3? Please note that the molar mass of phosphorus is 30.9 g/ mol and hydrogen is 1.008 g/mol.

Answer 1

Answer : The maximum number of grams of $PH_3$ formed is, 8.955 g

Solution : Given,

Mass of phosphorous = 8.2 g

Mass of hydrogen = 4 g

Molar mass of $P_4$ = 123.6 g/mole

Molar mass of $H_2$ = 2.016 g/mole

Molar mass of $PH_3$ = 33.924 g/mole

The balanced chemical reaction is,

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

First we have to calculate the moles $P_4$ and $H_2$

$\text{ Moles of }P_4=\frac{\text{ Mass of }P_4}{\text{ Molar mass of }P_4}=\frac{8.2g}{123.6g/mole}=0.066moles$

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{4g}{2.016g/mole}=1.98moles$

From the reaction, we conclude that

1 mole of $P_4$ react with 6 moles of $H_2$

0.066 moles of $P_4$ react with $6\times 0.066=0.396$ moles of $H_2$

That means the $H_2$ is in excess amount and $P_4$ is in limited amount.

Now we have to calculate the moles of $PH_3$.

As, 1 mole of $P_4$ react to give 4 moles of $PH_3$

So, 0.066 moles of $P_4$ react to give $4\times 0.066=0.264$ moles of $PH_3$

Now we have to calculate the mass of $PH_3$

$\text{ Mass of }PH_3=\text{ Moles of }PH_3\times \text{ Molar mass of }PH_3$

$\text{ Mass of }PH_3=(0.264moles)\times (33.924g/mole)=8.955g$

Therefore, the maximum number of grams of $PH_3$ formed is, 8.955 g