Question

Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction takes place

$Zn+2HCl\to ZnC{l}_2+H_2$

Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HCl. 1 mol of a gas occupies 22.7 L volume at STP; atomic mass of Zn = 65.3u

Answer 1

Given that, Mass of Zn = 3265 g

1 mole of gas occupies = 22.7L volum at STP Atomic mass of Zn = 65.3u

The given equation is

$\underset{65.3g}{Zn}+2HCl⟶ZnC{l}_2+\underset{1mol=22.7LatSTP}{H_2}$

From the above equatoin, it is clear that

65.3 g Zn, when reacts with HCl, produces = 22.7 L of $H_2$ at STP

$\therefore 32.65g$ Zn, when reacts with HCl, will produce $=\frac{22.7\times 32.65}{65.3}=11.35Lof{H}_2$ at STP.