Answer to Question #325764 in Chemistry for Lynn

Question

1. When 10g of N2 and 7g H2 are allowed to react. Determine the limiting reactant and excess reactant in this reaction.

N_{2}+H_{2} ->NH_{3}

Given:

Find:

Solution:

Answer 1

Given:

Mass of N₂ = 10 g

Mass of H₂ = 7 g

Find:

The limiting reactant is ???

The excess reactant is ???

Solution:

The molar mass of N₂ is 28.02 g/mol

The molar mass of H₂ is 2.016 g/mol

Calculate the moles of each reactant:

Moles of N₂ = (10 g N₂) × (1 mol N₂ / 28.02 g N₂) = 0.357 mol N₂

Moles of H₂ = (7 g H₂) × (1 mol H₂ / 2.016 g H₂) = 3.472 mol H₂

Balanced chemical equation:

N₂ + 3H₂ → 2NH₃

According to the chemical equation above:

1 mol of N₂ reacts with 3 mol of H₂

Thus, 0.357 mol of N₂ reacts with:

(0.357 mol N₂) × (3 mol H₂ / 1 mol N₂) = 1.071 mol H₂

However, initially there is 3.472 of H₂ (according to the task)

Therefore, N₂ acts as limiting reactant and H₂ is excess reactant

Answer:

The limiting reactant is N₂

The excess reactant is H₂