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 volume of nitrogen dioxide liberated in (b).
Topic: Mole concept, gas volumes and stoichiometry
Answer
Zn + 4HNO₃ ⟶ Zn(NO₃)₂ + 2H₂O + 2NO₂
From the equation, 1 mol Zn gives 2 mol NO₂.
Moles of Zn = 32.5 ÷ 65 = 0.5 mol Moles of NO₂ = 2 × 0.5 = 1 mol
Volume at STP = moles × 22.4 = 1 × 22.4 = 22.4 litres
Volume of nitrogen dioxide at STP = 22.4 litres
Go back to the zinc, not to the acid you calculated in part (b) — the zinc is the measured quantity and using it keeps the working short.
The coefficient 2 in front of NO₂ matters: 0.5 mol of zinc gives 1 mol of NO₂, not 0.5. Miss it and you get 11.2 litres. Then convert with 22.4 litres per mole, because the question asks for a volume at STP.