Question

The amount of zinc required to produce 1.12 ml of $H_2$ at STP on treatment with dilute HCl will be:

• A. $65g$
• B. $0.065g$
• C. $32.5\times {10}^{-4}g$
• D. $6.5g$

Answer 1

The correct option is C $32.5\times {10}^{-4}g$

When zinc reacts with dilute HCl, the formation of zinc chloride with hydrogen gas takes place.

$Zn+2HCl⟶ZnC{l}_2+H_2$

Therefore, $1$ mole of Zn$\equiv$$1$ mole of $H_2$

At STP, $22,400$ml of $H_2\equiv$ $1$ mole of $H_2$

$\therefore 1.12$ ml of $H_2\equiv 5\times {10}^{-5}$ mole of $H_2$

$\therefore$ Moles of Zn = $5\times {10}^{-5}$ mole

$\therefore$ Amount of Zn = $5\times {10}^{-5}\times 65$

= $32.5\times {10}^{-4}$ g