MCQs

O₂ is limiting. One mole NO (30 g) is formed.

0.1 mol NaOH is insufficient to react completely with 0.1 mol H₂SO₄.

Stoichiometric reaction produces 4 mol CO₂ = 89.6 dm³.

A is limiting. Product formed = (2/3) × 3 = 2 mol.

Industries optimize limiting reactants to maximize efficiency.

Stoichiometric mixture produces 3 mol S = 96 g.

Mg is limiting. Two moles Mg produce 2 mol MgO = 80 g.

The limiting reactant determines the theoretical yield.

O₂ limits the reaction. Only 2 mol P₄ react, leaving 2 mol P₄.

The reactants are present in the exact stoichiometric ratio.

It is the reactant consumed first, stopping the reaction.

CaCO₃ = 0.1 mol, HCl = 0.274 mol. Since only 0.2 mol HCl is required, CaCO₃ is limiting.

One mole of Fe₂O₃ reacts with 3 mol CO to produce 2 mol Fe = 112 g.

H₂ is limiting. NH₃ formed = (2/3) × 4 = 2.67 mol.

Al = 2 mol, Cl₂ = 1 mol. Two moles Al require 3 mol Cl₂, so chlorine is limiting.

The reaction is 1:1. Two moles of O₂ produce only 2 moles of CO₂.

8 mol O₂ require only 6.4 mol NH₃. Therefore, NH₃ remains in excess by 1.6 mol.

The ratio 10:15 simplifies to 2:3, matching the balanced equation. Neither reactant is in excess.

4 g H₂ = 2 mol and 32 g O₂ = 1 mol. The ratio is exactly 2:1, so both reactants are completely consumed.