The balanced equation is
#2"Al(s)" + "3Zn"("NO"_3)_2("aq") → 2"Al"("NO"_3)_3"(aq)" + 3"Zn(s)"#
One way is to replace the #"NO"_3# groups with #"X"# and then balance by inspection.
We recognize that groups within parentheses usually behave as a unit.
The unbalanced equation is
#"Al(s)" + "3Zn"("NO"_3)_2("aq") → "Al"("NO"_3)_3"(aq)" + "Zn(s)"#
Let's replace the #"NO"_3# groups by single symbols #"X"#.
Then the equation becomes
#"Al(s)" + "3ZnX"_2("aq") → "AlX"_3"(aq)" + "Zn(s)"#
Balance Zn
We have 3 #"Zn"# atoms on the left, so we need 3 #"Zn"# on the right. Put a #3# in front of #"Zn"#.
#"Al(s)" + "3ZnX"_2("aq") → "AlX"_3"(aq)" + color(red)(3)"Zn(s)"#
Balance #"X"#
We have 6 #"X"# on the left, so we need 6 #"X"# on the right. Put a #2# in front of #"AlX"_3#.
#"Al(s)" + "3ZnX"_2("aq") → color(blue)(2)"AlX"_3"(aq)" + color(red)(3)"Zn(s)"#
Balance #"Al"#
We have fixed 2 #"Al"# atoms on the right, so we need 2 #"Al"# atoms on the left. Put a #2# in front of #"Al"#
#color(fuchsia)(2)"Al(s)" + "3ZnX"_2("aq") → color(blue)(2)"AlX"_3"(aq)" + color(red)(3)"Zn(s)"#
The equation should now be balanced. Let's check.
On the left: #"2 Al, 3 Zn, 6 X"#
On the right: #"2 Al, 6 X, 3 Zn"#
Now we replace all the #"X"#s with #"NO"_3# and get the balanced equation.
#2"Al(s)" + "3Zn"("NO"_3)_2("aq") → 2"Al"("NO"_3)_3"(aq)" + 3"Zn(s)"#
