Question

Balance the following chemical equation:

Cu(s)+HNO3(aq)=Cu(NO3)2+NO(g)+H2O(l).

I missed this question on a quiz after trying to balance it both algebraically and by standard guess and check methods. Several classmates missed it as well. The answer is all over the internet, but what I need is a step by step explanation on how this would be balanced. Thank you!

Answer 1

Guess and check method:

Step 1: We can see there is 1 Cu on left and on 1 Cu on right side of the equation.

So let us consider the N on both sides. On left side of equation there is 1 N and on right side there are 3 N, so let us take 3 as coefficient of on the left side. So the N get balanced. The equation becomes

Step 2: Now there are 3 H on left and 2 on right, so let's take 6 as coefficient of and 3 as coefficient of (so on both sides H becomes 6 in number). The equation becomes:

Step 3: Now we can see that N is 6 on left and 3 on right. So if we take 2 as coefficient of Cu(NO₃)₂ and 2 as coefficient of NO, then N becomes 6 (2x2=4 in 2 molecules of copper nitrate and 2x1=2 in 2 molecules of NO) on both sides. Then the equation becomes:

Step 4: Now there are 2 Cu on right and 1 on left, so let's have 2 as coefficient of Cu on left. So Cu becomes 2 on both sides. The equation becomes

Step 5: i) Now we can see that number of O on left is 6x3=18 in HNO₃ whereas on right 2x3x2=12 O in copper nitrate and 2x1=2 in NO and 3x1=3 in water, that makes it 12+2+3=17 which means O is not balanced. So let's consider 8 as coefficient of Cu(NO ₃)₂ on left and 4 as coefficient of water on right. So on both sides of equation H becomes 8 and

ii) Now there are 8 N on left and 6 on right, so to balance that let's consider 3 as coefficient of Cu(NO₃)₂ (that means 3x2=6 N) and 2 as coefficient of NO. That makes it 8 N on both left and right side of equation. Also this way there are 8x3=24 O on left and from 3 molecules of copper nitrate 3x3x2=18, from 2 molecules of NO there are 2x1=2 O and from 4 molecules of water there are 4x1=4 O, so total of 18+2+4=24 O in on right side of equation. So N, H and O are balanced on both sides of the equation given:

Step 6: Now we have 2 Cu on left and 3 on right, so we can simply change the coefficient of Cu on left to 3 and the equation becomes:

So we can see that atoms of all elements are now balanced

Cu=3 on both sides

H=8 on both sides

O=24 on both sides

N=8 on both sides