A mixture containing 28\text{ g} of nitrogen gas (N_2) and 12\text{ g} of hydrogen gas (H_… — JAMB UTME

A mixture containing

28\text{ g}

of nitrogen gas (

N_2

) and

12\text{ g}

of hydrogen gas (

H_2

) reacts to completion according to the equation:

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

What mass of the excess reactant remains unreacted at the end of the reaction?

[N = 14,\ H = 1]

The mass of the excess reactant remaining unreacted is

6.0\text{ g}

The correct answer is

6.0\text{ g}

28\text{ g}

N_2

corresponds to

1.0\text{ mol}

, while

12\text{ g}

H_2

corresponds to

6.0\text{ mol}

. According to the equation

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

1.0\text{ mol}

N_2

consumes

3.0\text{ mol}

H_2

. Therefore,

N_2

limits the reaction, and

3.0\text{ mol}

H_2

(

6.0\text{ g}

) remains unreacted.

Calculate the initial number of moles for each reactant.

\text{Moles of } N_2 = \frac{28\text{ g}}{28\text{ g/mol}} = 1.0\text{ mol}

;

\text{Moles of } H_2 = \frac{12\text{ g}}{2\text{ g/mol}} = 6.0\text{ mol}

Converting given masses to moles is necessary to apply stoichiometric coefficients.

Determine the limiting reactant using the mole ratio from the balanced equation.

The reaction ratio is

1\text{ mol } N_2 : 3\text{ mol } H_2

1.0\text{ mol } N_2

requires

3.0\text{ mol } H_2

. Since

6.0\text{ mol } H_2

is available,

N_2

is the limiting reactant and

H_2

is in excess.

Comparing available mole ratios against stoichiometric requirements identifies which reactant is completely consumed.

Calculate the unreacted moles and mass of the excess reactant (

H_2

\text{Unreacted moles of } H_2 = 6.0\text{ mol} - 3.0\text{ mol} = 3.0\text{ mol}

\text{Unreacted mass} = 3.0\text{ mol} \times 2\text{ g/mol} = 6.0\text{ g}

Multiplying unreacted moles by the molar mass of

H_2

yields the remaining mass.

Limiting and Excess Reactants in Stoichiometry

Estimated Time:1m 30s

Question