COLLEGE CHEMISTRY · STOICHIOMETRY

Who runs out first?

Adjust the reactant amounts, predict what limits the reaction, and see how much product the available mixture can make.

N₂(g)+3 H₂(g)→2 NH₃(g)

Balanced synthesis of ammonia · coefficients give the mole ratio 1 : 3 : 2

01 · SET THE MIXTURE

Starting reactants

Amount in moles

Nitrogen, N₂2 mol

0 mol5 mol10 mol

Enter amount mol

Hydrogen, H₂4 mol

0 mol5 mol10 mol

Enter amount mol

Try a mixture

02 · FOLLOW THE STOICHIOMETRY

LIMITING REACTANT · H₂ (hydrogen)

Reactants consumed in a 1 : 3 ratio2.67 mol NH₃ possible

THEORETICAL YIELD45.42 g

NH₃(g), based on the limiting reactant

AMMONIA PRODUCED2.67 mol

2 moles NH₃ per reaction unit

N₂ LEFT OVER0.67 mol

Excess nitrogen remaining

H₂ LEFT OVER0 mol

Excess hydrogen remaining

Show the calculation

1. Compare possible reaction units. Divide each reactant amount by its coefficient:

N₂: 2 mol ÷ 1 = 1.33 reaction units
H₂: 4 mol ÷ 3 = 1.33 reaction units

2. Use the smaller value. 1.33 reaction units can occur, so the limiting reactant is h₂ (hydrogen).

3. Convert to product. 1.33 reaction units × 2 mol NH₃ per unit = 2.67 mol NH₃.

4. Convert moles to mass. 2.67 mol NH₃ × 17.031 g/mol = 45.42 g NH₃ theoretical yield.

Assumes the reaction goes to completion with no side reactions or product loss. In an actual experiment, measured yield may be lower.