The FIRST problem is that the loss of electrons for N2 is 4. N is +4 for the 2 N atoms and it goes to zero for the 2N atoms on the right. EACH N is -2 but you aren't dealing with EACH N. You are dealing with both on the left and both on the right.
Second thing, not enough to start World War III over, but you should not say that Br changes by a factor of 6. It changes by 6 electrons, yes, and its a GAIN of 6e but not a FACTOR of 6.
Br gains 6 electrons, both N change by 4 electrons. Multiply N by 6 and Br by 4 and see if things get easier.
Hello, I am having trouble balancing this redox reaction WITHOUT using the half-reaction method. I believe this is because not all of the elements are in (aq).
BrO3-(aq) + N2H4(g) --> Br-(aq) + N2(g)
Br is reduced by a factor of 6.
N is oxidized by a factor of 2.
I will include these factors as coefficients and attempt to balance the hydrogen and oxygen.
2BrO3-(aq) + 6N2H4(g) --> 2Br-(aq) + 6N2(g)+ 6H2O(l)
It seems like I should be adding 12H+ to the products side to balance the hydrogen, but this doesn't work. I know the answer is:
2BrO3-(aq) + 3N2H4(g) --> 2Br-(aq) + 3N2(g)+ 6H2O(l)
But why are my stoichiometric factors incorrect for N2H4 and N2?
Thank you!
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