Consider the following reactions: (i)Cu+HNO3(dil.)→ (ii)Cu+HNO3(conc.)→ (iii)Zn+HNO3(20%)→ (iv)Zn+HNO3(60%)→
Total number of reactions in which N2 is released is/are:
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Solution
The corresponding reactions are: 3Cu+8HNO3dilute→2NO+3Cu(NO3)2+4H2O Cu+4HNO3Conc.→2NO2+Cu(NO3)2+2H2O 4Zn+10HNO3(20%)→4Zn(NO3)2+N2O+5H2O Zn+4HNO3(60%)→Zn(NO3)2+2NO2+2H2O
So, total number of reactions in which N2 is released is 0.