$Z n + 2 H C l ⟶ Z n {C l}_{2} + H_{2}$
$H_{2}$ gas produced by the reaction is collected over water. If 0.3 g of Zn are consumed at $27^{0}$ in the reaction, calculate the volume of hydrogen produced. (Given: Barometric reading=750mm, Aqueous tension at $27$ =25 mm, Atomic weight of Zn=65.4).

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Solution

1 mole $Z n \to$ 1 mole $H_{2}$
$\frac{65.4 g m}{218} Z n \to \frac{1 m o l e}{218} H_{2}$
0.3 gm $Z n \to \frac{1}{48}$ mole $H_{2}$
$\Rightarrow$ 1 mole $H_{2} \to 22.4 L$
$\frac{1}{218}$ mole $H_{2} \to \frac{22.4}{218} L = 0.10275$ litres.

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Zn+2HCl⟶ZnCl2+H2H2 gas produced by the reaction is collected over water. If 0.3 g of Zn are consumed at 270 in the reaction, calculate the volume of hydrogen produced. (Given: Barometric reading=750mm, Aqueous tension at 27 =25 mm, Atomic weight of Zn=65.4).

1 mole Zn→ 1 mole H265.4gm218Zn→1mole218H20.3 gm Zn→148 mole H2⇒ 1 mole H2→22.4L1218 mole H2→22.4218L=0.10275 litres.

$Z n + 2 H C l ⟶ Z n {C l}_{2} + H_{2}$
$H_{2}$ gas produced by the reaction is collected over water. If 0.3 g of Zn are consumed at $27^{0}$ in the reaction, calculate the volume of hydrogen produced. (Given: Barometric reading=750mm, Aqueous tension at $27$ =25 mm, Atomic weight of Zn=65.4).

1 mole $Z n \to$ 1 mole $H_{2}$
$\frac{65.4 g m}{218} Z n \to \frac{1 m o l e}{218} H_{2}$
0.3 gm $Z n \to \frac{1}{48}$ mole $H_{2}$
$\Rightarrow$ 1 mole $H_{2} \to 22.4 L$
$\frac{1}{218}$ mole $H_{2} \to \frac{22.4}{218} L = 0.10275$ litres.