Question

How can I balance this Chemistry equation? `Cu+HNO_3 -> Cu(NO_3)_2 + NO_2+ H_2O`

Answer 1

`Cu+4HNO_3->Cu(NO_3)_2+2NO_2+2H_2O`

Explanation:

We have the unbalanced chemical reaction:

`Cu+HNO_3->Cu(NO_3)_2+NO_2+H_2O`

Let's expand out `(NO_3)_2` using the distributive property from algebra, so it'll be easier to work with.

`Cu+HNO_3->CuN_2O_6+NO_2+H_2O`

I see that there are two hydrogen atoms on the right side, while only one on the left side, so let's multiply hydrogen by two.

`Cu+2HNO_3->CuN_2O_6+NO_2+2H_2O`

This is the tricky part here, which you might not be able to get. By inspection, I see that if I multiply `NO_2` by `2`, I will have a total of `4` nitrogen atoms on the right side, and if I also multiply the `HNO_3` by `2`, I will also have `4` nitrogens on the left side.

Note that this method takes some practice, before you can get used to it.

`=>Cu+4HNO_3->CuN_2O_6+2NO_2+2H_2O`

Now, I see that I have `4*3=12` oxygens on the left, and `6+2*2+2*1=12` oxygens on the right, so they are balanced. There are `4` hydrogens on the left, and `2*2=4` hydrogens on the right. There are `4` nitrogens on the left, and `2+2*1=4` on the right. Copper is still the same on both sides, just like in the unbalanced equation.

Finally, we can put back `N_2O_6` into `(NO_3)_2`, like the ion it is, and our final equation is:

`color(blue)(barul(|Cu+4HNO_3->Cu(NO_3)_2+2NO_2+2H_2O|)`

Answer 2

It's a tough one. I've worked through my thinking process below. In the end, I got stuck. I hope another answerer can see the way from here.

Explanation:

Initially there is one `Cu` on the left and one on the right, but there is one `H` on the left and two on the right, and similarly one `NO_3` on the left and two on the right.

`Cu+HNO_3 -> Cu(NO_3)_2 + NO_2+ H_2O`

It might be better to look at `N` and `O` separately rather than at `NO_3`, because without doing so we have ignored the `NO_2`.

Initially there is one `N` on the left and three on the right, and three `O` on the left and nine on the right.

Let's try taking three `HNO_3`:

`Cu+3` `HNO_3 -> Cu(NO_3)_2 + NO_2+ 3/2` `H_2O`

I believe this is balanced:

`Cu:` `1` left, `1` right
`H:` `3` left, `3`(as `3/2xx2`) right
`N:` `3` left, `3` right
`O:` `9` left, `9 1/2` right

Dang! So close!

We could multiply through by `2` if the fractions are uncomfortable (they shouldn't be, since the equation represents moles and half a mole is a thing), but it doesn't solve this issue.

Let's just do that anyway, see if it helps us think:

`2` `Cu+6` `HNO_3 -> 2` `Cu(NO_3)_2 + 2` `NO_2+ 3` `H_2O`

`Cu:` `2` left, `2` right
`H:` `6` left, `6` right
`N:` `6` left, `6` right
`O:` `18` left, `19` right

Still got that same problem.

I have to admit to being stuck at this point. I'll ask other answerers to check and see whether they can clarify.