6.54 g of zinc are treated with 11.5 g of H_2SO_4. Calculate the volume of hydrogen evolved at S.T.P. How much H_2SO_4 will be left in excess?

6.54 g of zinc are treated with 11.5 g of H_2SO_4. Calculate the volum

1.	6.54 g of zinc are treated with 11.5 g of H_2SO_4. Calculate the volume of hydrogen evolved at S.T.P. How much H_2SO_4 will be left in excess?
Answer» Solution :The CORRESPONDING CHEMICAL equation is: `underset(65.38 g)(Zn) + underset(2.016 + 32.06 + 64.0= 98.1 g)(H_(2)SO_(4)) to ZnSO_(4) + underset(22.4 "L at S.T.P")(H_(2))` From the equation, it is clear that 6.54 g of zinc will react with 9.81 g of `H_2SO_4`. Thus, the amount of `H_2SO_4` TAKEN is in excess. Hence, Zn will get consumed completely and this is the limiting reagent. `therefore 65.38 g` of zinc EVOLVE `H_(2) = 22.4 L` at S.T.P. `therefore 6.54 g` of zinc will evolve `H_(2) = 22.4/(65.38) xx 6.54 = 2.24` L at S.T.P. Further, `therefore 65.38` g of zinc react with `=H_(2)SO_(4) = 98.1 g` `therefore H_(2)SO_(4)` left in excess `=11.5- 98.1 = 1.69 g` Hence, 2.24 L hydrogen will be evolved at S.T.R and 1.69 g of `H_2SO_4` will be left in excess.