Balance the following chemical equation: C3H5N3O9 → CO2 + H2O + N2 + O2

There are 3 carbon atoms on the reactant side in $\text{C}_3\text{H}_5\text{N}_3\text{O}_9$. To balance this, place a coefficient of 3 in front of $\text{CO}_2$ on the product side.

$$\text{C}_3\text{H}_5\text{N}_3\text{O}_9 \rightarrow 3\text{CO}_2 + \text{H}_2\text{O} + \text{N}_2 + \text{O}_2$$

There are 5 hydrogen atoms on the reactant side in $\text{C}_3\text{H}_5\text{N}_3\text{O}_9$. To balance this, place a coefficient of $\frac{5}{2}$ in front of $\text{H}_2\text{O}$ on the product side.

$$\text{C}_3\text{H}_5\text{N}_3\text{O}_9 \rightarrow 3\text{CO}_2 + \frac{5}{2}\text{H}_2\text{O} + \text{N}_2 + \text{O}_2$$

There are 3 nitrogen atoms on the reactant side in $\text{C}_3\text{H}_5\text{N}_3\text{O}_9$. To balance this, place a coefficient of $\frac{3}{2}$ in front of $\text{N}_2$ on the product side.

$$\text{C}_3\text{H}_5\text{N}_3\text{O}_9 \rightarrow 3\text{CO}_2 + \frac{5}{2}\text{H}_2\text{O} + \frac{3}{2}\text{N}_2 + \text{O}_2$$

Count oxygen atoms on both sides:
Reactant side: 9 O atoms in $\text{C}_3\text{H}_5\text{N}_3\text{O}_9$.
Product side: $(3 \times 2) \text{ O in } 3\text{CO}_2 + (\frac{5}{2} \times 1) \text{ O in } \frac{5}{2}\text{H}_2\text{O} + (x \times 2) \text{ O in } x\text{O}_2$.
So, $9 = 6 + \frac{5}{2} + 2x$.
$9 = \frac{12}{2} + \frac{5}{2} + 2x$
$9 = \frac{17}{2} + 2x$
$18 = 17 + 4x$ (multiplying by 2 to clear the fraction)
$1 = 4x$
$x = \frac{1}{4}$
Place a coefficient of $\frac{1}{4}$ in front of $\text{O}_2$.

$$\text{C}_3\text{H}_5\text{N}_3\text{O}_9 \rightarrow 3\text{CO}_2 + \frac{5}{2}\text{H}_2\text{O} + \frac{3}{2}\text{N}_2 + \frac{1}{4}\text{O}_2$$

$$4(\text{C}_3\text{H}_5\text{N}_3\text{O}_9) \rightarrow 4(3\text{CO}_2) + 4(\frac{5}{2}\text{H}_2\text{O}) + 4(\frac{3}{2}\text{N}_2) + 4(\frac{1}{4}\text{O}_2)$$ $$4\text{C}_3\text{H}_5\text{N}_3\text{O}_9 \rightarrow 12\text{CO}_2 + 10\text{H}_2\text{O} + 6\text{N}_2 + \text{O}_2$$

Reactants: C = 12, H = 20, N = 12, O = 36
Products: C = 12, H = 20, N = 12, O = $(12 \times 2) + (10 \times 1) + (1 \times 2) = 24 + 10 + 2 = 36$
All atoms are balanced.

The balanced chemical equation is:

$$\boxed{4\text{C}_3\text{H}_5\text{N}_3\text{O}_9 \rightarrow 12\text{CO}_2 + 10\text{H}_2\text{O} + 6\text{N}_2 + \text{O}_2}$$