MnO₂ + 4HCl ⟶ MnCl₂ + 2H₂O + Cl₂

0.02 moles of pure MnO₂ is heated strongly with conc. HCl. Calculate:

(a) mass of MnO₂ used

(b) moles of salt formed,

(c) mass of salt formed,

(d) moles of chlorine gas formed,

(e) mass of chlorine gas formed,

(f) volume of chlorine gas formed at S.T.P.,

(g) moles of acid required,

(h) Mass of acid required.

$\begin{matrix} \text{MnO}$2 & + & 4\text{HCl}& \longrightarrow & \text{MnCl}2 & + & 2\text{H}2\text{O} & + & \text{Cl}2 \ 1 \text{ mole} && 4 \text{ mole} && 1\text{ mole}&&&& 1\text{ mole} \ 55 +2(16) && 4[1 + 35.5] & & 55 + 2(35.5) & & &&2(35.5) \ = 87 \text{ g} & & = 146 \text{ g} & & = 126 \text{ g} & & & & = 71\text{ g} \ \end{matrix}

(a) 1 mole of MnO₂ weighs 87 g

∴ 0.02 mole will weigh $\dfrac{87}{1}$ x 0.02 = 1.74 g

(b) 1 mole MnO₂ gives 1 mole of MnCl₂

∴ 0.02 mole MnO₂ will give 0.02 mole of MnCl₂

(c) As, 1 mole MnCl₂ weighs 126 g

∴ 0.02 mole MnCl₂ will weigh $\dfrac{126}{1}$ x 0.02 = 2.52 g

(d) 1 mole MnO₂ gives 1 mole of Cl₂

∴ 0.02 mole MnO₂will form 0.02 moles of Cl₂

(e) 1 mole of Cl₂ weighs 71 g

∴ 0.02 mole will weigh $\dfrac{71}{1}$ x 0.02 = 1.42 g

(f) 1 mole of chlorine gas has volume 22.4 dm³

∴ 0.02 mole will have volume $\dfrac{22.4}{1}$ x 0.02 = 0.448 dm³

(g) 1 mole MnO₂ requires 4 moles of HCl

∴ 0.02 mole MnO₂ will require $\dfrac{4}{1}$ x 0.02 = 0.08 mole

(e) Mass of 1 mole of HCl = 36.5 g

∴ Mass of 0.08 mole = 0.08 × 36.5 = 2.92 g