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Concentration Terms - An Introduction (w/w etc)

The minimum w...

Question

The minimum weight of $M n O_{2}$ and minimum volume of $H C l$ of specific gravity $1.2$ and $3.65 %$ by weight, needed to produce $1.12$ litres of $C l_{2}$ at STP by the reaction given as follows:

$M n O_{2} + H C l \to M n C l_{2} + H_{2} O + C l_{2}$

4.35 g, 200 m l

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48.7 g, 166.67 m l

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4.35 g, 166.7 m l

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4.35 g, 333.3 m l

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Solution

The correct option is B $4.35 g, 166.7 m l$
The number of moles of chlorine obtained $= \frac{1.12}{22.4} = 0.05$ moles.

They are also equal to the number of moles of manganese dioxide.

Mass of manganese dioxide $= 0.05 \times 86.9 = 4.35$ g.

The number of moles of $H C l$ $= 4 \times 0.05 = 0.2$ moles.

Mass of $H C l$ $= 0.2 \times 36.5 \times \frac{100}{3.65} = 200$ g.

Volume of $H C l$ $= \frac{200}{1.2} = 166.7$ ml.

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