Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc, Following reaction takes place
`Zn+2HClrarrZnCl_(2)+H_(2)`
Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HCl. 1 mole of a gas occupies 22.7 L volume at STP, atomic mass of Zn=65 .3u

Explanation

Given that, mass of Zn=32.65g
1 mole of gas occupies =22.7L volume at STP
Atomic mass of Zn=65.3u
The given equation is
`Zn+2HClrarrZnCl_(2)+H_(2)`
`65.3g 1"mol"=22.7L` at STP
From the above equation it is clear that
`65.3g Zn`. when reacts with HCl produces `=22.7L "of" H_(2)` at STP
`therefore 32.65gZn`. when reacts with HCl, will produce `=(22.7xx32.65)/(65.3)=11.35L` of `H_(2)` at STP.