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

Apr 3, 2018

$C u + 4 H N {O}_{3} \to C u {\left(N {O}_{3}\right)}_{2} + 2 N {O}_{2} + 2 {H}_{2} O$

#### Explanation:

We have the unbalanced chemical reaction:

$C u + H N {O}_{3} \to C u {\left(N {O}_{3}\right)}_{2} + N {O}_{2} + {H}_{2} O$

Let's expand out ${\left(N {O}_{3}\right)}_{2}$ using the distributive property from algebra, so it'll be easier to work with.

$C u + H N {O}_{3} \to C u {N}_{2} {O}_{6} + N {O}_{2} + {H}_{2} O$

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.

$C u + 2 H N {O}_{3} \to C u {N}_{2} {O}_{6} + N {O}_{2} + 2 {H}_{2} O$

This is the tricky part here, which you might not be able to get. By inspection, I see that if I multiply $N {O}_{2}$ by $2$, I will have a total of $4$ nitrogen atoms on the right side, and if I also multiply the $H N {O}_{3}$ 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.

$\implies C u + 4 H N {O}_{3} \to C u {N}_{2} {O}_{6} + 2 N {O}_{2} + 2 {H}_{2} O$

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

Finally, we can put back ${N}_{2} {O}_{6}$ into ${\left(N {O}_{3}\right)}_{2}$, like the ion it is, and our final equation is:

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

Apr 3, 2018

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 $C u$ on the left and one on the right, but there is one $H$ on the left and two on the right, and similarly one $N {O}_{3}$ on the left and two on the right.

$C u + H N {O}_{3} \to C u {\left(N {O}_{3}\right)}_{2} + N {O}_{2} + {H}_{2} O$

It might be better to look at $N$ and $O$ separately rather than at $N {O}_{3}$, because without doing so we have ignored the $N {O}_{2}$.

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 $H N {O}_{3}$:

$C u + 3$ $H N {O}_{3} \to C u {\left(N {O}_{3}\right)}_{2} + N {O}_{2} + \frac{3}{2}$ ${H}_{2} O$

I believe this is balanced:

$C u :$ $1$ left, $1$ right
$H :$ $3$ left, $3$(as $\frac{3}{2} \times 2$) right
$N :$ $3$ left, $3$ right
$O :$ $9$ left, $9 \frac{1}{2}$ 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:

$2$ $C u + 6$ $H N {O}_{3} \to 2$ $C u {\left(N {O}_{3}\right)}_{2} + 2$ $N {O}_{2} + 3$ ${H}_{2} O$

$C u :$ $2$ 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.