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Stoichiometric Calculations

MnO2+4HCl→ 2H...

Question

$M n O_{2} + 4 H C l \to 2 H_{2} O + M n {C l}_{2} + {C l}_{2}$
The reaction of $M n O_{2}$ and $H C l$ produces $2.24 L$ of ${C l}_{2}$ gas. What mass of $M n O_{2}$ was used up by the reaction? (At. mass of $M n = 55 g / m o l$ )

7.19 g

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8.7 g

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2.24 g

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18.69

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Solution

The correct option is D $8.7 g$
$M n O_{2} + 4 H C l \to M n C l_{2} + C l_{2} + 2 H_{2} O$

$1 m o l e$ of $M n O_{2}$ produces $22.4 L i t r e s$ of $C l_{2}$ in the reaction at STP.

$1 m o l e$ $M n O_{2} \to$ $22.4 L C l_{2}$

The molar mass of $M n O_{2} = 55 + 2 \times 16 = 87 g$

$1 m o l e$ of $M n O_{2} = 87 g$

$2.24 L C l_{2}$ $= \frac{(2.24) (87)}{22.4} = 8.7 g o f M n O_{2}$

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