Step 1: Products are $CO$ and $H_2O$.

$C_3H_8 + O_2 \rightarrow CO + H_2O$

Step 2: Balance C. Need 3 CO.

$C_3H_8 + O_2 \rightarrow \mathbf{3}CO + H_2O$

Step 3: Balance H. Need 4 H₂O.

$C_3H_8 + O_2 \rightarrow 3CO + \mathbf{4}H_2O$

Step 4: Balance O. Right: $(3 \times 1) + (4 \times 1) = 7$. Need 7 on Left. $\rightarrow 3.5 \: O_2$.

Multiply mainly by 2 to clear fractions:

$2C_3H_8 + 7O_2 \rightarrow 6CO + 8H_2O$