C-H,03 + C, H,03 → C, H,O, + H2O
Balance the above equation.​

C
3
H
8
+
5
O
2
→
3
C
O
2
+
4
H
2
O

Explanation:
We have:
C
3
H
8
+
O
2
→
C
O
2
+
H
2
O

I would always start by balancing the elements that only occur once on each side. In this case it's carbon. We also notice that as it now stands we have an odd number of oxygen molecules on the right and an even number on the left, which won't work. We can fix this by multiplying the water by an even number.
C
3
H
8
+
O
2
→
C
O
2
+
4
H
2
O

Notice I multiplied it by
4
because we want to have 8 hydrogens on both sides.
Now I multiply the
C
O
2
by
3
to match the amount of
C
as on the LHS.
C
3
H
8
+
O
2
→
3
C
O
2
+
4
H
2
O

The last step is to determine the coefficient of the oxygen gas on the LHS. We see that we have
6
+
4
=
10
oxygen molecules on the RHS, so we will have
10
2
=
5
molecules on the LHS. The final balanced chemical equation is
C
3
H
8
+
5
O
2
→
3
C
O
2
+
4
H
2
O

This is a perfect example of a combustion reaction because we have a carbon based compound reaction with oxygen gas to produce carbon dioxide and water.
Hopefully this helps!