Question

Hydrogen gas is prepared in the laboratory by reacting dilute \[\ce{HCl}\] with granulated zinc. Following reaction takes place.

\[\ce{Zn + 2HCl -> ZnCl2 + H2}\]

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

Answer 1

Given that, mass of \[\ce{Zn}\] = 32.65 g

1 mole of gas occupies = 22.7 L volume at STP Atomic mass of \[\ce{Zn}\] = 65.3u

The given equation is

\[\ce{\underset{65.3g}{Zn} + 2HCl -> ZnCl2 + \underset{1 mol = 22.7L at STP}{H2}}\]

From the above equation, it is clear that 65.3 g of \[\ce{Zn}\] when reacts with \[\ce{HCl}\] produces = 22.7 L \[\ce{H2}\] at STP

∴ 32.65 g of \[\ce{Zn}\] when reacts with HCl will produce = `(22.7 xx 32.65/65.3` = 11.35 L of \[\ce{H2}\] at STP.