Two masses? Do not start with the product yet.

For the reaction N₂ + 3H₂ → 2NH₃, 28 g of N₂ reacts with 9.0 g of H₂.
What mass of NH₃ is produced? (N = 14, H = 1)

A) 17 g
B) 34 g
C) 51 g
D) 76.5 g

The question gives you amounts of two reactants. That is your signal to check the limiting reactant before calculating the product.

Do not assume that both reactants will be used completely. If they are present in exactly the ratio shown by the balanced equation, both can be consumed together. If they are not, one runs out first and limits the amount of product.

Step 1 — Convert both masses to moles.

• N₂: molar mass = 2 × 14 = 28 g/mol
  28 ÷ 28 = 1 mol
• H₂: molar mass = 2 × 1 = 2 g/mol
  9.0 ÷ 2 = 4.5 mol

Step 2 — Read the ratio from the balanced equation.

N₂ + 3H₂ → 2NH₃

The equation tells you:

1 mol N₂ needs 3 mol H₂

Step 3 — Find the limiting reactant.

You have 1 mol of N₂. To react completely, it needs 3 mol of H₂.

You actually have 4.5 mol of H₂, so there is more hydrogen than required. The nitrogen runs out first.

Therefore:

N₂ is the limiting reactant.

After 1 mol of N₂ reacts, 1.5 mol of H₂ remains in excess.

Step 4 — Calculate the product from the limiting reactant.

From the balanced equation:

1 mol N₂ → 2 mol NH₃

So 1 mol of N₂ produces:

2 mol NH₃

Step 5 — Convert ammonia to grams.

Molar mass of NH₃ = 14 + (3 × 1) = 17 g/mol

2 × 17 = 34 g

Answer: B — 34 g.

Where does option C come from?

If you calculate from all 4.5 mol of H₂, you get:

4.5 ÷ 3 = 1.5
1.5 × 2 = 3 mol NH₃
3 × 17 = 51 g

The arithmetic is correct, but it starts from the excess reactant. Producing 3 mol of NH₃ would require 1.5 mol of N₂, while the question gives only 1 mol.

That is why 51 g is the trap.

The ten-second check

When a question gives amounts of two reactants, check the limiting reactant first:

1. Convert both amounts to moles.
2. Divide each mole amount by its coefficient in the balanced equation.
3. The smaller result identifies the limiting reactant.

Here:

• N₂: 1 ÷ 1 = 1
• H₂: 4.5 ÷ 3 = 1.5

Since 1 is smaller, N₂ is limiting. If the two results were equal, the reactants would be present in the exact required ratio.