Question

What is the answer to this question?

Answer 1

Well, you have to balance mass and balance charge....mate...

Explanation:

And of course you could do this by trial and error....with each of the proferred solutions. Alternatively you could do this from first principles, using redox equations

Copper metal is oxidized to `"copper(II) ion"`

...`Cu(s) rarr Cu^(2+)+2e^(-)` `(i)`

And the nitrogen in nitrate, `N(+V)`, is REDUCED to `N(+II)`, and `N(+IV)`, in `NO`, and `NO_2` respectively....

`2NO_3^(-)+6H^+ +4e^(-) rarrNO(g) + NO_2(g)+3H_2O(l)` `(ii)`

And so we adds `(i)` and `(ii)` in a linear combination such that the electrons are removed as virtual particles....i.e. `2xx(i)+(ii)`

`2Cu(s) +2NO_3^(-)+6H^+ +4e^(-)rarr 2Cu^(2+)+NO(g) + NO_2(g)+3H_2O(l)+4e^(-)`

...to give after cancellation....

`2Cu(s) +2NO_3^(-)+6H^+ rarr 2Cu^(2+)+NO(g) + NO_2(g)+3H_2O(l)`

...and we could tart this equation up a bit by adding `4xxNO_3^(-)` to each side....

`2Cu(s) +6HNO_3(aq) rarr 2Cu(NO_3)_2(aq)+NO(g) + NO_2(g)+3H_2O(l)`

...the which is balanced with respect to mass and charge, and corresponds to `"option 1"`....