The amount of zinc required to produce 1.12mL of H2 at STP on treatment with dil.HCl will be: (Given: Molar mass of Zn= 65g/mol)
A
60g
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B
0.65g
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C
32.5×10−4g
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D
6.5g
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Solution
The correct option is C32.5×10−4g The reaction of Zinc with dil.HCl gives: Zn+2HCl→ZnCl2+H2 22400mL of H2 is produced from 65g of zinc.
So, 1.12mL of H2 will be produced from (6522400×1.12)=32.5×10−4g of zinc.