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Stoichiometric Calculations

The weight of...

Question

The weight of $M n O_{2}$ and the volume of $H C l$ of specific gravity $1.2$ g/mL and $4 %$ nature by weight needed to produce $1.78$ L of $C l_{2}$ at $S T P$ . The reaction involved is as follows:
$M n O_{2} + 4 H C l \to M n C l_{2} + 2 H_{2} O + C l_{2}$

A

True

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B

False

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Solution

The correct option is B False
$M n O_{2} + 4 H C l \to M n C l_{2} + 2 H_{2} O + C l_{2}$
Moles $0.08$ $4 \times 0.8$ $\frac{1.78}{22.4} = 0.08$
$= 0.32$
$\Rightarrow W_{M n O_{2}} = 0.08 \times 87 g = 6.96 g;$
$(\frac{10 \times 4 \times 1.2}{36.5}) \times V_{L} = 0.32 \Rightarrow V_{L} = 0.24 L$

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Similar questions

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H C l

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M n O_{2} + 4 H C l \to M n C l_{2} + 3 H_{2} O + C l_{2}

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