Redox and Equivalent ConceptHard

Question

Aqueous HCl reacts with ${MnO}_{2}(\text{ }s)$ to form ${MnCl}_{2}(aq)$, ${Cl}_{2}(\text{ }g)$ and $H_{2}O(l)$. What is the weight (in g ) of ${Cl}_{2}$ liberated when 8.7 g of ${MnO}_{2}(\text{ }s)$ is reacted with excess aqueous HCl solution? (Given Molar mass in $g{mol}^{- 1}Mn = 55,Cl = 35.5,O = 16,H = 1$ )

Options

A.7.1
B.71
C.21.3
D.14.2

Solution

${MnO}_{2} + 4HCl \rightarrow {MnCl}_{2} + {Cl}_{2} + 2H_{2}O$

$\frac{8.7}{87}\ $ Excess

$= 0.1$ mole 0.1 mole

Wt. of ${Cl}_{2}$ obtained $= 0.1 \times 71 = 7.1\text{ }g$

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