Question

\[\ce{Zn + 4HNO3 -> Zn(NO3)2 + 2H2O + 2NO2}\]

32.5 g of zinc reacts with concentrated nitric acid as given in the above equation.

1. How many moles of zinc was required in the reaction?
2. Find the mass of nitric acid needed to react with 32.5 g of zinc.
3.   Find the volume of nitrogen dioxide liberated in (b).

[Atomic weight: H = 1, N = 14, O = 16, Zn = 65]

Answer 1

\[\ce{Zn + 4HNO3 -> Zn(NO3)2 + 2H2O + 2NO2}\]

Given: mass of Zn = 32.5 g

Molar mass of Zn = 65 g/mol

(a) Moles of Zn = `32.5/5`

= 0.5 mol

(b) 1 mol of Zn reacts with 4 moles of HNO₃.

moles of HNO₃ = 0.5 × 4 

= 2.0 mol

Molar mass HNO₃ = 1(H) + 14(N) + 3 × 16(O)

= 63 g mol⁻¹

mass(HNO₃) = 2.0 mol × 63 g mol⁻¹ 

= 126 g

(c) 1 mol Zn produces 2 moles of NO₂.

moles of NO₂ = 0.5 × 2

= 1.0 mol

At STP, 1 mol of any gas at STP occupies 22.4 L.

Volume of NO₂ = 1.0 × 22.4 L 

= 22.4 L