6.5g of $Z n$ is dissolved in excess of $H_{2} {S O}_{4}$ . The weight of ${Z n S O}_{4}$ formed and the volume of $H_{2}$ gas liberated at $S T P$ are $(Z n = 65)$

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$161 g, 2.24 L$

$16.1 g, 2.24 L$

$16.1 g, 22.4 L$

$16.1 g, 11.2 L$

answer is B.

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Detailed Solution

No. of moles of $Z n$ = 6.5/65= 0.1 mole and $H_{2} {S O}_{4}$ is in excess

$Z n + H_{2} {S O}_{4} \to {Z n S O}_{4} + H_{2} 65 g 161 g 22.4 l 6.5 g ? ?$

The mass of ZnSO₄ formed= $\frac{6.5}{65} \times 161 = 16.1 g$

Volume of H₂ liberated= $\frac{6.5}{65} \times 22.4 = 2.24 l$

Hence option (B) is the correct Answer.

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