How much zinc should be treated with excess of dilute hydrochloric acid to obtain 2.24 litres of hydrogen at S.T.P.

How much zinc should be treated with excess of dilute hydrochloric aci

1.	How much zinc should be treated with excess of dilute hydrochloric acid to obtain 2.24 litres of hydrogen at S.T.P.
Answer» Solution :The balanced chemical equation representing the combination of zinc with hydrochloric acid is: `underset(1 "mole 1 gram atom 65.38 g")(ZN) + 2HCL (aq) to ZnCl_(2)(aq) + underset("1 mole 22.4 L at S.T.P")(H_(2)(g))` This equation signifies that 1 mole (1 gram atom), i.e., 65.38 g of zinc react with hydrochloric acid to produce 1 mole, i.e., 22.4 L of hydrogen at S.T.P `therefore 22.4 L` of hydrogen at S.T.R are GIVEN by zinc = 65.38 g `therefore 2.24` L of hydrogen at S.T.P will be given by zinc `=(65.38)/22.4 xx 2.24 = 6.54 g` Hence, 2.24 L hydrogen at ST. R will be PRODUCED by the action of 6.54 g of zinc with excess of HCl.