Mole Concept and Stoichiometry — Question 10

$\begin{matrix} \text{MnO}_2 & + & 4\text{HCl}& \longrightarrow & \text{MnCl}_2 & + & 2\text{H}_2\text{O} & + & \text{Cl}_2 \\ 1 \text{ mole} && 4 \text{ mole} && 1\text{ mole}&&&& 1\text{ mole} \\ 55 +2(16) && 4[1 + 35.5] & & 55 + 2(35.5) & & &&2(35.5) \\ = 87 \text{ g} & & = 146 \text{ g} & & = 126 \text{ g} & & & & = 71\text{ g} \\ \end{matrix}$

(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