Question

Ammonia is formed according to the reaction below. A chemist mixes 21 grams of nitrogen gas and 18 grams of hydrogen gas in a 2.0 L vessel. How many grams of hydrogen gas will be consumed?

Answer 1

Answer : The mass of hydrogen gas consumed will be, 4.5 grams

Explanation : Given,

Mass of
`N_2` = 21 g

Mass of
`H_2` = 18 g

Molar mass of
`N_2` = 28 g/mole

Molar mass of
`H_2` = 2 g/mole

First we have to calculate the moles of
`N_2` and
`H_2`.

`\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=(21g)/(28g/mole)=0.75moles`

`\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=(18g)/(2g/mole)=9moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`N_2(g)+3H_2(g)\rightarrow 2NH_3(g)`

From the given balanced reaction, we conclude that

As, 1 mole of
`N_2` react with 3 moles of
`H_2`

So, 0.75 moles of
`N_2` react with
`3* 0.75=2.25` moles of
`H_2`

From this we conclude that,
`H_2` is an excess reagent because the given moles are greater than the required moles and
`N_2` is a limiting reagent because it limits the formation of product.

The moles of hydrogen gas consumed = 2.25 mole

Now we have to calculate the mass of hydrogen gas consumed.

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

`\text{Mass of }H_2=(2.25mole)* (2g/mole)=4.5g`

Therefore, the mass of hydrogen gas consumed will be, 4.5 grams