Zn + 4HNO3 ⟶ Zn(NO3)2 + 2H2O + 2NO2
32.5 g of zinc reacts with concentrated nitric acid as given in the above equation.
(a) How many moles of zinc was required in the reaction?
(b) Find the mass of nitric acid needed to react with 32.5 g of zinc.
(c) Find the volume of nitrogen dioxide liberated in (b).
[Atomic weight: H=1, N=14, O=16, Zn=65]
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$$\begin{matrix} \text{Zn} & + & 4\text{HNO}_3 & \longrightarrow \\ 1 \text{ mole} && 4 \text{ moles} & \\ 65 && 4[1 + 14 + 3(16)] & \\ = 65 \text{ g} & & = 252 \text{ g} & \\ & & & \\ \text{Zn(NO}_3)_2 & + & 2\text{H}_2\text{O} & + & 2\text{NO}_2 \\ 1\text{ mole}&& 2 \text{ moles} && 2\text{ moles} \\ 65 + 2[14 + 3(16)] & & 2[2(1) + 16] && 2[14 + 2(16)] \\ = 189 \text{ g} & & = 36 \text{ g} & & = 92 \text{ g} \\ \end{matrix}$$
(a) Number of moles of Zn = $\frac{\text{Mass}}{\text{Molar mass}}$ = $\frac{32.5}{65}$ = 0.5 mole
(b) From the equation, 65 g of Zn reacts with 252 g of HNO3
∴ 32.5 g of Zn reacts with $\frac{252}{65}$ × 32.5 = 126 g of HNO3
(c) From the equation, 65 g of Zn produces 2 × 22.4 = 44.8 L of NO2 at STP
∴ 32.5 g of Zn produces $\frac{44.8}{65}$ × 32.5 = 22.4 L of NO2 at STP
Hence, 0.5 mole of zinc is required, 126 g of nitric acid is needed, and 22.4 L of NO2 is liberated at STP.