Identify the limiting reactant in the reaction of nitrogen and hydrogen to form NH3 if 5.23 g of N2 and 5.52 g of H2 are combined. Determine the amount (in grams) of excess reactant that remains after the reaction is complete.

1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

2) The amount (in grams) of excess reactant H₂ = 4.39 g.

Explanation:

Firstly, we should write the balanced equation of the reaction:

N₂ + 3H₂ → 2NH₃.

1) To determine the limiting reactant of the reaction:

From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃. This means that N₂ reacts with H₂ with a ratio of (1:3). We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation: n = mass / molar mass.

The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.

The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.

From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).

The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).

∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

2) To determine the amount (in grams) of excess reactant of the reaction:

As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂. Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess. The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol. ∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol) ((2.015 g/mol) = 4.39 g.