i am completely baffled by this equation- CAN SOMEONE PLEASE FIND A WAY TO BALANCE IT?
Cu + HNO3 ----> Cu(NO3)2 + NO +H2O
Cu + HNO3 ----> Cu(NO3)2 + NO +H2O
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3 Cu + 8 HNO3 ----> 3 Cu(NO3)2 + 2 NO + 4 H2O
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This is quite popular reaction of HNO3
Apply oxidation charge:
Cu donors 2e-, N receives 3e-
--> 3Cu + HNO3 ----> 3Cu(NO3)2 + 2NO +H2O
Since from the right side having 8N:
---> 3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO +H2O
Now, left side has 8H:
--> 3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O
Use O to double check: 24O
Apply oxidation charge:
Cu donors 2e-, N receives 3e-
--> 3Cu + HNO3 ----> 3Cu(NO3)2 + 2NO +H2O
Since from the right side having 8N:
---> 3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO +H2O
Now, left side has 8H:
--> 3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O
Use O to double check: 24O
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3Cu + 8HNO3 ---> 3Cu(NO3)2 + 2NO + 4H2O
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the equation does not seem to be correct