A mixture of aluminium and zinc weighing 1.67 g was completely dissolved in acid and 1.69 litres of hydrogen measured at 0°C and 1 atmospheric pressure were evolved. What was the original weight of aluminium in the mixture?

A mixture of aluminium and zinc weighing 1.67 g was completely dissolv

1.	A mixture of aluminium and zinc weighing 1.67 g was completely dissolved in acid and 1.69 litres of hydrogen measured at 0°C and 1 atmospheric pressure were evolved. What was the original weight of aluminium in the mixture?
Answer» Solution :Both aluminium and zinc react with sulphuric acid to form hydrogen. Suppose, the mixture contains x g of aluminium. HENCE, amount of zinc in the mixture = `1.67 -x g`. The corresponding equations are: `underset(53.96 g)(2Al) + H_(2)SO_(4) to underset(3 xx 22.4 "L at S.T.P.")(Al_(2)(SO_(4))_(3)) + 3H_(2)` `underset(65.38 g)(Zn) + H_(2)SO_(4) to underset(22.4 L "at S.T.P")(ZnSO_(4)) + H_(2)` `therefore 53.96` g of Al evolve `H_(2) = 3 xx 22.4` L at S.T.P. `therefore x GM` of Al will evolve `H_(2) = (3 xx 22.4)/(53.96) xx x` `=1.24 x` L at S.T.P. SIMILARLY, `therefore 1.24 x + (22.4)/(65.38) xx (1.67 -x) = 1.69` which gives `x = 1.24` Therefore, the WEIGHT of aluminium in the given sample is 1.24 g.