Given a balanced Which reactant is the limiting reactant and what is the mass (in grams) remaining of the reactant in excess?

Question

Given a balanced chemical equation,

   N₂(g) + 3 H₂(g) ⟶ 2 NH₃(g)

If a reaction mixture contains 2 moles of N₂(g) and 7 moles of H₂(g), which reactant is the limiting reactant and what is the mass (in grams) remaining of the reactant in excess? 

Answer 1

To determine the limiting reactant, we need to compare the stoichiometry of the reactants and see which one is completely consumed first. The balanced equation for the reaction between nitrogen gas (N2) and hydrogen gas (H2) to form ammonia (NH3) is:

N2(g) + 3H2(g) -> 2NH3(g)

According to the balanced equation, the molar ratio between N2 and H2 is 1:3.

Given that the reaction mixture contains 2 moles of N2 and 7 moles of H2, we can calculate the moles of NH3 that can be produced from each reactant:

For N2: Moles of NH3 produced = (2 moles N2) × (2 moles NH3 / 1 mole N2) = 4 moles NH3

For H2: Moles of NH3 produced = (7 moles H2) × (2 moles NH3 / 3 moles H2) ≈ 4.67 moles NH3

From the calculations, we can see that N2 can produce 4 moles of NH3, while H2 can produce approximately 4.67 moles of NH3. Since the reaction cannot produce fractional moles, N2 is the limiting reactant because it produces the lesser amount of NH3.

To find the mass remaining of the reactant in excess (H2), we need to calculate the moles of H2 that are in excess, which is the difference between the moles of H2 present in the reaction mixture and the moles of H2 required to react with the limiting reactant (N2).

Moles of H2 in excess = (moles of H2 in the reaction mixture) - (moles of H2 required to react with N2) = 7 moles - (4 moles NH3 × (3 moles H2 / 2 moles NH3)) ≈ 7 moles - 6 moles = 1 mole H2

Now, to find the mass of H2 remaining in grams, we need to use its molar mass. The molar mass of hydrogen gas (H2) is approximately 2 g/mol.

Mass of H2 remaining = (moles of H2 in excess) × (molar mass of H2) = 1 mole × 2 g/mol = 2 grams

Therefore, the mass remaining of the reactant in excess (H2) is 2 grams.