**Zn + 2HCl → ZnCl _{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 = 65.3 u.**

Calculation:

1 mole of gas occupies = 22.7l volume at STP

Atomic mass of Zinc (Zn) = 65.3u

65.3g of Zn when reacts with HCl produces 22.7L H_{2} at STP

Hence,

32.65g of Zinc when reacts with hydrochloric acid will produce 11.35L of H_{2} at STP.