The amount of zinc required to produce 1.12 ml of H 2 at STP on treatment with dilute HCl will be

Question

The amount of zinc required to produce 1.12 ml of H 2 at STP on treatment with dilute HCl will be (A) 65 g (B) 0.065 g (C) 32.5 × 10-4 g (D) 6.5 g

Tardigrade Admin · Accepted Answer

underset1 moLZn +2 HCl longrightarrow underset22.4 L = 22400 mlZnCl 2+ H 2( g ) 224000 ml of H 2 gas is produced from Zn =65 g 1.12 ml of H 2 gas is produced from Zn =(65/22400) × 1.12 g =3.25 × 10-3 g i.e., 32.5 × 10-4 g

Q. The amount of zinc required to produce $1.12 m l$ of $H_{2}$ at STP on treatment with dilute $H Cl$ will be

$65 g$

$0.065 g$

$32.5 \times 1 0^{- 4} g$

$6.5 g$

$1 m o L Z n + 2 H Cl ⟶ 22.4 L = 22400 m l Z n C l_{2} + H_{2} (g)$
$224000 m l$ of $H_{2}$ gas is produced from $Z n = 65 g$
$1.12 m l$ of $H_{2}$ gas is produced from
$Z n = \frac{65}{22400} \times 1.12 g$
$= 3.25 \times 1 0^{- 3} g$
i.e., $32.5 \times 1 0^{- 4} g$

Q. The amount of zinc required to produce 1.12ml of H2​ at STP on treatment with dilute HCl will be

65g

0.065g

32.5×10−4g

6.5g