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Variation of Pressure in Fluids

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Question

Calculate the weight of $M n O_{2}$ .
$H C l$ of specific gravity $1.2$ g/mL and $5 %$ by weight needed to produce $1.12 L$ of $C l_{2}$ at STP by the reaction (as nearest integer).
$M n O_{2} + 4 H C l \to M n C l_{2} + 3 H_{2} O + C l_{2}$

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Solution

(a) $N_{H C l} = \frac{% b y w e i g h t \times 10 \times d}{E w o f H C l}$
$= \frac{5 \times 10 \times 1.2}{36.5} = 1.64$
Now, mEq of $M n O_{2} =$ mEq of $H C l$
=mEq of $C l_{2}$ formed.
$= \frac{1.12}{11.2} \times 10^{3}$
$= 100$ $[\begin{matrix} E w & o f & C l_{2} & = \frac{M}{2} 1 & e q & o f & C l_{2} & = 11.2 L \end{matrix}]$
(b) Volume of $H C l$ used:
$N \times V = 100$
$1.64 \times V = 100$
$∴ V_{H C l} = 60.97 m L$
Because $H C l$ is also used to give $M n C l_{2}$ thus volume used is double than required for the reduction of $M n O_{2}$ .
$∴ V_{H C l} = 2 \times 60.97 = 121.94 m L$
(c) Also, mEq of $M n O_{2} = m E q$ of $H C l = 100$
$\frac{W}{87 / 2} \times 10^{3} = 100 (E w o f M n O_{2} = \frac{55 + 32}{2})$
$∴ W_{M n O_{2}} = 4.35 g$
nearest integer value is $4 g m$ .

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