The amount of zinc (at. wt. = 65) necessary to produce 224 ml of H2 by the reaction with an acid will be

Question

The amount of zinc (at. wt. = 65) necessary to produce 224 ml of H2 by the reaction with an acid will be (A) 6.5 gm (B) 0.065 gm (C) 7.5 gm (D) 0.65 gm

Tardigrade Admin · Accepted Answer

Zn + H2SO4 → ZnSO4 + H2 65 g Zn will give 22.4 litre of H2 at STP ⇒ 224 ml of H2 corresponds to (65/22400)×224 = 0.65 g Zn

Q. The amount of zinc (at. wt. = 65) necessary to produce $224 m l$ of $H_{2}$ by the reaction with an acid will be

6.5 gm

0.065 gm

7.5 gm

0.65 gm

$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$

$65 g$ Zn will give 22.4 litre of $H_{2}$ at STP

$\Rightarrow 224 m l$ of $H_{2}$ corresponds to $\frac{65}{22400} \times 224$

$= 0.65 g Z n$

Q. The amount of zinc (at. wt. = 65) necessary to produce 224ml of H2​ by the reaction with an acid will be