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Mole Concept and Avogadro's Number

calculate the...

Question

calculate the mass of $M n O_{2}$ , needed to produce 1.78 litres of chlorine gas at STP according to the equation
$M n O_{2}$ + 4HCl $\to$ $M n C l_{2}$ + 2 $H_{2} O$ + $C l_{2}$

[H = 1, Cl = 35.5, Mn = 54.9]

6.905g

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25.2 ml

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6.1g

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25.9 ml

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Solution

The correct option is A
6.905g

The equation is
$M n O_{2}$ + 4HCl $\to$ $M n C l_{2}$ + $2 H_{2} O$ + $C l_{2}$
1 mole of $M n O_{2}$ = 1 mole of $C l_{2}$
86.9g of $M n O_{2}$ = 22.4 L of $C l_{2}$ at NTP
$∴$ 1.78 L $C l_{2}$ = $\frac{1.78 \times 86.9}{22.4}$ = 6.905 g of $M n O_{2}$

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