We are given the balanced chemical equation:
Zn + 4HNO3 → Zn(NO3)2 + 2H2O + 2NO2
Mass of zinc given = 32.5 g
Atomic masses:
H = 1, N = 14, O = 16, Zn = 65
We will solve each part step by step.
(a) Number of moles of zinc required in the reaction
First, find the molar mass of zinc:
Molar mass of Zn = 65 g/mol
Number of moles of zinc:
\[
\text{Moles of Zn} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{32.5}{65}
\]
\[
\text{Moles of Zn} = 0.5
\]
Therefore, the number of moles of zinc required is 0.5 mole.
(b) Mass of nitric acid needed to react with 32.5 g of zinc
From the balanced equation:
1 mole of Zn reacts with 4 moles of HNO3
Therefore,
0.5 mole of Zn will react with:
\[
0.5 \times 4 = 2 \text{ moles of HNO}_3
\]
Now calculate the molar mass of nitric acid, HNO3:
HNO3 = 1 + 14 + (3 × 16)
\[
= 1 + 14 + 48 = 63 \text{ g/mol}
\]
So, mass of 2 moles of HNO3:
\[
2 \times 63 = 126 \text{ g}
\]
Therefore, the mass of nitric acid needed is 126 g.
(c) Volume of nitrogen dioxide liberated
From the balanced equation:
1 mole of Zn gives 2 moles of NO2
Therefore,
0.5 mole of Zn gives:
\[
0.5 \times 2 = 1 \text{ mole of NO}_2
\]
At STP, 1 mole of any gas occupies 22.4 litres.
So, 1 mole of NO2 will occupy:
\[
22.4 \text{ litres}
\]
Therefore, the volume of nitrogen dioxide liberated is 22.4 litres.
Final Answers:
(a) Number of moles of zinc = 0.5 mole
(b) Mass of nitric acid required = 126 g
(c) Volume of nitrogen dioxide liberated = 22.4 litres
Quick Check:
- 32.5 g Zn = 32.5 ÷ 65 = 0.5 mol
- HNO3 needed = 4 × 0.5 = 2 mol
- Mass of HNO3 = 2 × 63 = 126 g
- NO2 formed = 2 × 0.5 = 1 mol
- Volume of 1 mol gas at STP = 22.4 L