Question

how many moles of h2 gas will evolve when 32.5g of zinc reacts completely with dilute h2so4​

Answer 1

Step-by-step explanation:

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

Answer 2

Answer:

0.5mole

Step-by-step explanation:

the balanced equation for the given reaction is

Zn+H2SO4----->ZnSO4+H2

zn atomic weight is 65g so when 1 mole of zn reacts 1 mole of hydrogen is formed. zn is completely used so it is limiting reagent.so evolved hydrogen will depend on zn.

But given weight is 32.5g means 1/2 mole

so when 1/2 mole of zn reacts then 1/2 mole of hydrogen will be evolved.

THEREFORE YOUR ANSWER IS 1/2=0.5 MOLE.

I hope you are satisfied by my answer.