Question

If 1.0 g of H2 reacts with excess N2 according to the reaction below, how much NH3 will be formed?
3H2(g) + N2(g) ? 2NH3(g)

Answer 1

Moles of H2 = Mass / Molar Mass

=> Moles = 1 / 2 = 0.5 moles

Given that N2 is in excess quantity.

Hence H2 is in limited quantity and the stoichiometry will be governed by H2

The reaction is,

3H2 + N2 ----> 2NH3

According to the reaction,

3 moles of H2 will react with 1 mole of N2 to produce 2 moles of NH3

Therefore, 0.5 moles of H2 will react with 0.5/3 = 0.167 moles of N2 to produce 2 x 0.5 / 3 = 0.333 moles of NH3

Molar mass of NH3 = 17

Moles of NH3 produced = 0.333 moles

Mass of NH3 produced = 0.333 x 17 = 5.667 g