Question

If 62.7 g N2 react with 23.8 g H2 according to the reaction below, how many grams of ammonia (NH3) can be produced, and how many grams of the excess reactant will be left over?

Unbalanced equation: N2 + H2 → NH3

Answer 1

10.3648 g of hydrogen gas will be left over.

Explanation:

Balanced chemical equation:

`N_2 + 3H_2\rightarrow 2NH_3`

Moles of nitrogen gas =
`(62.7 g)/(28 g/mol)=2.2392 mol`

Moles of hydrogen gas =
`(23.8 g)/(2 g/mol)=11.9 mol`

According to reaction 1 mol of nitrogen gas reacts with 3 mol of hydrogen gas.Then 2.2392 moles of nitrogen gas will react with:

`(3)/(1)* 2.2392 mol=6.7176 mol` of hydrogen

Mass of 6.7176 moles of hydrogen gas = 6.7176 mol × 2 g/mol = 13.4352 g

Nitrogen gas is in limited amount which will limit the formation of ammonia gas.

Mass of hydrogen gas left unreacted = 23.8 g - 13.4352 g=10.3648 g

10.3648 g of hydrogen gas will be left over.

Answer 2

1. The balanced chemical reaction is:

N2 + 3H2 → 2NH3

We are given the amounts of the reactants to be used in the reaction. These will be the starting point for the calculations.

62.7 g N2 ( 1mol / 28.02 g ) = 2.24 mol N2
23.8 g H2 ( 1 mol / 2.02 g) = 11.78 mol H2

The molar ratio of the reactants is 1 is to 3. Therefore, the limiting reactant is nitrogen. To completely react the 2.24 mol N2 only 6.72 mol of H2 is needed.

6.72 mol H2 excess ( 2.02 g / 1 mol ) = 13.57 g excess