Question

In the industrial preparation of ammonia, nitrogen and hydrogen gases are are reacted according to the reaction N2(g) + 3H2(g) 2NH3(g). Suppose 466 kg of N2 are mixed with 39.4 kg of H2

How many moles of N2(g) and of H2(g) will remain when the reaction is complete?

Answer 1

At first, the limiting reactant should be identified. For the purpose, the number of moles of reactants must be estimated.

Mr(N₂) = 28 g/mol

Mr(H₂) = 2 g/mol

Next, according to the reaction, the number of moles of the reactants are: n(N₂) = 466 × 10³ g / 28 g/mol = 16.64 × 10³ mol and n(H₂) = 39.4 × 10³ g / 6 g/mol = 6.57 × 10³ mol. As a result, hydrogen gas is a limiting reactant and nitrogen gas is in excess. The reaction occurs when equal amounts of both gasses react. As a result, 16.64 × 10³ - 6.57 × 10³ = 10.07 × 10³ moles of nitrogen will not react and will remain while all amount of hydrogen will be used in the reaction.

Answer: 10.07 × 10³ mol of N₂(g) and 0 mol of H₂(g).