Question

What is the maximum mass in grams of NH3 that can be produced by the reaction of of 2.5 g N2 with 2.5 g of H2 via the equation below?

N2 (g) + 3 H2 (g) → 2 NH3 (g)

Answer 1

Answer: The mass of
`NH_3` produced is, 3.03 grams.

Explanation : Given,

Mass of
`N_2` = 2.5 g

Mass of
`H_2` = 2.5 g

Molar mass of
`N_2` = 28 g/mol

Molar mass of
`H_2` = 2 g/mol

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

`\text{Moles of }N_2=\frac{\text{Given mass }N_2}{\text{Molar mass }N_2}=(2.5g)/(28g/mol)=0.089mol`

and,

`\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}=(2.5g)/(2g/mol)=1.25mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

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

From the balanced reaction we conclude that

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

So, 0.089 moles of
`N_2` react with
`0.089* 3=0.267` 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 and it limits the formation of product.

Now we have to calculate the moles of
`NH_3`

From the reaction, we conclude that

As, 1 mole of
`N_2` react to give 2 mole of
`NH_3`

So, 0.089 mole of
`N_2` react to give
`0.089* 2=0.178` mole of
`NH_3`

Now we have to calculate the mass of
`NH_3`

`\text{ Mass of }NH_3=\text{ Moles of }NH_3* \text{ Molar mass of }NH_3`

Molar mass of
`NH_3` = 17 g/mole

`\text{ Mass of }NH_3=(0.178moles)* (17g/mole)=3.03g`

Therefore, the mass of
`NH_3` produced is, 3.03 grams.