Question

When 28 g of nitrogen and 6 g of hydrogen react, 34 g of ammonia are produced. if 100 g of nitrogen react with 6 g of hydrogen, how much ammonia will be produced?

Answer 1

N2 + 3H2 --> 2NH3
When 100g of N2 , no of moles of N2= 100/(28)=3.57 mol
no. of moles of h2 = 6/(2)=3mol
Therefore h2 is limiting reagent.
no. of moles of ammonia= 3/3*2=2moles
mass of ammonia produced= 2 mol * (14+3)= 34g

Answer 2

Answer : The mass of
`NH_3` produced will be, 34 grams.

Explanation : Given,

Mass of
`N_2` = 100 g

Mass of
`H_2` = 6 g

Molar mass of
`N_2` = 28 g/mole

Molar mass of
`H_2` = 2 g/mole

Molar mass of
`NH_3` = 17 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}=(100g)/(28g/mole)=3.57moles`

`\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=(6g)/(2g/mole)=3moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`N_2+3H_2\rightarrow 2NH_3`

From the balanced reaction we conclude that

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

So, the given 3 moles of
`N_2` react with 1 moles of
`H_2`

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

Now we have to calculate the moles of
`NH_3`.

As, 3 moles of
`H_2` react to give 2 moles of
`NH_3`

So, 3 moles of
`H_2` react to give
`(3)/(3)* 2=2` moles 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`

`\text{Mass of }NH_3=(2mole)* (17g/mole)=34g`

Therefore, the mass of
`NH_3` produced will be, 34 grams.