Question

What volume of hydrogen measured at ${27}^{\circ }C$ and 776.7 mm pressure would be obtained by treating 0.65 g of pure zinc with sufficient amount of dilute hydrochloric acid? Aq. tension at ${27}^{\circ }C$ is 26.7 mm; At. mass of Zn is 65.

• A. 200.43 ml
• B. 300.43 ml
• C. 249.43 ml
• D. 244.43 ml

Answer 1

The correct option is C

249.43 ml

The reaction of Zn replacing $H_2$ from $H_{2}S{O}_4$.
$\therefore$ Zn + $H_{2}S{O}_4$ $\to$ $H_{2}S{O}_4$ + $H_2$

$\therefore$ 65g of Zn = 22.4 L of $H_2$ at NTP

0.65g of Zn = 0.224 L of $H_2$ at NTP = 224 ml
Other pressure and temperature are given and volume at that condition is asked.

Thus,use $\frac{P_1{V}_1}{T_1}$ = $\frac{P_2{V}_2}{T_2}$
$P_2=776.7-$ aqueous tension

$\Rightarrow$ $\frac{760\times 224}{273}$ = $\frac{750\times V_2}{300}$

$\Rightarrow$ $V_2$ = 249.43 ml