Chlorine is prepared in the laboratory by treating manganese dioxide ( $M O_{2})$ with aqueous hydrochloric acid according to the reaction,
$M n O_{2} + 4 H C l \to M n C l_{2} + C l_{2} + 2 H_{2} O$
How many grams Via will react with $5 g M n O_{2}$ ?

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Solution

A balanced equation is:
$M n O_{2} + 4 H C I \to M n C l_{2} + C l_{2} + 2 H_{2} O$
According to the given equation:
1 mole of $M n O_{2}$ reacts with 4 mole of $H C l$
or 87 g $M n O_{2}$ reacts with 146 g $H C l$
$∴$ 5 g $M n O_{2}$ will react with $\frac{146}{87} \times 5 = 8.39 g H C l$

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