Balance the chemical equation: C3H8 + O2 ightarrow CO2 + H2O

There are 3 carbon atoms on the left side ($C_{3}H_{8}$). To balance this, place a coefficient of 3 in front of $CO_{2}$ on the right side.

$$C_{3}H_{8} + O_{2} \rightarrow 3CO_{2} + H_{2}O$$

There are 8 hydrogen atoms on the left side ($C_{3}H_{8}$). To balance this, place a coefficient of 4 in front of $H_{2}O$ on the right side (since $4 \times 2 = 8$).

$$C_{3}H_{8} + O_{2} \rightarrow 3CO_{2} + 4H_{2}O$$

Count the oxygen atoms on the right side:
From $3CO_{2}$: $3 \times 2 = 6$ oxygen atoms.
From $4H_{2}O$: $4 \times 1 = 4$ oxygen atoms.
Total oxygen atoms on the right = $6 + 4 = 10$.
There are 2 oxygen atoms on the left side ($O_{2}$). To balance this, place a coefficient of 5 in front of $O_{2}$ (since $5 \times 2 = 10$).

$$C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O$$

Reactants:
C: 3
H: 8
O: $5 \times 2 = 10$

Products:
C: $3 \times 1 = 3$
H: $4 \times 2 = 8$
O: $(3 \times 2) + (4 \times 1) = 6 + 4 = 10$

All atoms are balanced.