Chlorine is prepared in the laboratory by treating manganese dioxide ( $M n O_{2}$ ) with aqueous hydrochloric acid according to the reaction.

$4 H C l (a q) + M n O_{2} (s) \to 2 H_{2} O (l) + M n C l_{2} (a q) + C l_{2} (g)$ .

How many grams of $H C l$ react with 5.0 g of manganese dioxide?

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Solution

$5.0 g$ of $M n O_{2}$ contains= $\frac{5.0 g}{87 m o l^{- 1} g} = 0.0574$ moles

$4 H C l (a q) + M n O_{2} (s) ⟶ 2 H_{2} O + M n C l_{2} + C l_{2} (g)$

$1$ mole of $M n O_{2}$ reacts with $4$ mole of $H C l$

Thus, $0.0574$ mole of $M n O_{2}$ will react with $4 \times 0.0574$ moles of $H C l$

$= 0.23$ mole of $H C l$ will react with $5 g$ of $M n O_{2}$

Amount of $H C l$ in $0.23$ mole= $0.23 \times 36.5$ = $8.4 g$ of $H C l$

Thus, $8.4 g$ of $H C l$ will react with $5 g$ of $M n O_{2}$ .

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