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

2 Answers
Apr 3, 2018

Answer:

#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|)#

Apr 3, 2018

Answer:

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.