Question

The amount of zinc required to produce 224 ml of
H, at STP on treatment with dilute H,SO, will be
(Zn = 65)
(1) 65 g
(2) 0.065 g
(3) 0.65 g
(4) 6.5 g​

Answer 1

Answer:

0.65g

Explanation:

Zn+H2SO4=ZnSO4+H2

Now,

To produce 22400ml H2, we need=65 g Zn

Therefore,to produce 224 ml H2, we need ={(65÷22400)×224}g Zn=0.65 g Zn