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}=(8.2g)/(123.6g/mole)=0.066moles`

`\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=(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* 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* 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* \text{ Molar mass of }PH_3`

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

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