Question

The volume of NH3 gas up in one 150 ML of H2 gas is treated with hundred ML of N2

Answer 1

Assume that reaction took place at standard conditions

We know that, 22.4 L = 1 mole

so,

150 mL H2 = 1*0.15/22.4 = 0.0066 moles of H2

100 ml N2 = 0.1/22.4 = 0.0044 moles of N2

We know that,

N2 + 3H2  ---------> 2NH3

so, 1 mole of N2 reacts with 3 moles of H2

so, 0.0044 moles will react with 0.0132

Hence, H2 is limiting reagent.

3 mol of H2 form 2 mol. of NH3

so, 0.0044 mol of H2 will form = 0.0657 mol of NH3

hence, volume of NH3 = 65 mL