Guru Prasad Singh
Last Activity: 12 Years ago
The given reactions are as follows:
2KClO3------> 2KCl+3O2
2H2+O2----> 2H2O
Zn+H2SO4-----> ZnSO4+H2
Now, 122.5 of KClO3 gives 96 g of O2
Therefore O2 obtained from 24.5 gm of KClO3= 96/122.5*24.5= 19.2 gm of O2
Now 32 gm of O2 react with 4gm of H2 to give 36 gm of H2O.
Therefore Mass of H2 reacting with 19.2 gm of O2=4/32*19.2= 2.4gm
Now, 64 gm of Zn gives 2 gm of H2 gas.
Therefor Amount of Zn required to produce 2.4 gm of H2= 64/2*2.4=76.8 gm
Hence Amount of Zinc required for the given condition of reaction is 76.8gm.
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