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Q. The weight of $MnO _2$ and the volume of $HCl$ of specific gravity $1.2 \,g \, mL ^{-1}$ and $4 \%$ nature by weight, needed to produce $1.78 \, L$ of $Cl _2$ at $STP$. The reaction involved is:
$MnO _4+4 \,HCl \rightarrow MnCl _2+2 H _2 O + Cl _2$

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Solution:

$MnO _2+4 HCl \rightarrow MnCl _2+2 H _2 O + Cl _2$
Moles $0.08 4 \times 0.8 \frac{1.78}{22.4}=0.08$
$=0.32$
$\Rightarrow W_{ MnO _2}=0.08 \times 87\, g =6.96 \,g$
$\left(\frac{10 \times 4 \times 1.2}{36.5}\right) \times V_{ L }=0.32 \Rightarrow V_{ L }=0.24 \,L$