Question

10gram of impure zinc reacts with excess of dilute sulphuric acid to yield zinc sulphate and hydrogen. Calculate the moles of sulphuric acid consumed, mass of zinc sulphate formed and volume of hydrogen evolved at NTP.

Answer 1

10 gram of impure zinc reacts with excess of dilute sulphuric acid to yield zinc sulphate and hydrogen.

Zn(s) + H₂SO₄(aq) →  ZnSO₄(aq) + H₂(g)↑

In this reaction 

1 mole of Zn = 65 g 

but it is given that zinc is 10 g 

we can say that 

65 g Zn used to form 161 g  of ZnSO4 

but 10 g used to form (161 /65) x 10 = 24.6g 

also 1 mole of Sulphuric acid  = 98 g 

[Note :  65 g of Zn used with 98 g of H2SO4 
also 10 g of Zn used with (98 x 2) / 13 = 7.5 x 2 = 15.0g ]

And 1 g of Sulphuric acid = 1/98 moles 

15 g of sulphuric acid       = 15/98 moles = 0.53 Moles 

Volume of H2 at STP = moles of H2 formed X 22.4 

Amount of H2 formed (by calculation) = 2 x 0.15 = 0.3 g 

1 mole of H2 = 2 g 

1 g = 1/2 moles 

0.3 g = 0.3 x 1/2 = 0.15 moles 

So volume of H2 at STP = 0.15 x 22.4 = 3.36 L