Zn + 4HNO₃ → Zn(NO₃)₂ + 2H₂O + 2NO₂ 32.5 g of zinc reacts with concentrated nitric acid as given in the above equation. [Atomic weight: H=1, N=14, O=16, Zn=65]
Find the mass of nitric acid needed to react with 32.5 g of zinc.
Topic: Mole concept, gas volumes and stoichiometry
Answer
Zn + 4HNO₃ ⟶ Zn(NO₃)₂ + 2H₂O + 2NO₂
[H = 1, N = 14, O = 16, Zn = 65]
Molar masses: Zn = 65 g mol⁻¹ HNO₃ = 1 + 14 + (3 × 16) = 1 + 14 + 48 = 63 g mol⁻¹
From the equation, 1 mol Zn reacts with 4 mol HNO₃: 65 g of Zn reacts with 4 × 63 = 252 g of HNO₃
For 32.5 g of Zn (which is half of 65 g): mass of HNO₃ = 252 ÷ 2 = 126 g
Mass of nitric acid needed = 126 g
Spot that 32.5 g is exactly half of 65 g and the whole calculation halves with it — no long division needed.
The coefficient 4 is the trap. It is 4 × 63 = 252 g of acid per 65 g of zinc, not 63. Using 63 gives 31.5 g, a quarter of the right answer.