Q. \( \text{c}_3\text{h}_8 + \text{o}_2 \rightarrow \text{co}_2 + \text{h}_2\text{o} \) is a balanced chemical equation.

We want to balance the chemical equation:

\[
\mathrm{C_3H_8 + O_2 \rightarrow CO_2 + H_2O}
\]

Balancing means making sure the number of each type of atom on the reactant side equals the number on the product side.

Step 1: Assign coefficients.

Let the balanced equation be written as:

\[
a\,\mathrm{C_3H_8 + b\,O_2 \rightarrow c\,CO_2 + d\,H_2O}
\]

Here \(a\), \(b\), \(c\), and \(d\) are coefficients we will choose. We can usually take \(a = 1\) for simplicity because equations can be scaled.

So we start with:

\[
\mathrm{C_3H_8 + b\,O_2 \rightarrow c\,CO_2 + d\,H_2O}
\]

Step 2: Balance carbon atoms.

On the left, \(\mathrm{C_3H_8}\) contains \(3\) carbon atoms.

On the right, \(\mathrm{CO_2}\) contains \(1\) carbon per molecule, and there are \(c\) molecules of \(\mathrm{CO_2}\).

So carbon balance gives:

\[
3 = c
\]

Therefore:

\[
c = 3
\]

Step 3: Balance hydrogen atoms.

On the left, \(\mathrm{C_3H_8}\) contains \(8\) hydrogen atoms.

On the right, \(\mathrm{H_2O}\) contains \(2\) hydrogen atoms per molecule, and there are \(d\) molecules.

So hydrogen balance gives:

\[
8 = 2d
\]

Therefore:

\[
d = 4
\]

Step 4: Balance oxygen atoms.

Now count oxygen atoms.

Left side oxygen: \(b\) molecules of \(\mathrm{O_2}\) gives:

\[
2b
\]

Right side oxygen: \(\mathrm{CO_2}\) has \(2\) oxygen atoms each, and there are \(c = 3\) of them, so that contributes:

\[
3 \times 2 = 6
\]

\(\mathrm{H_2O}\) has \(1\) oxygen atom each, and there are \(d = 4\) of them, so that contributes:

\[
4 \times 1 = 4
\]

Total oxygen on the right is:

\[
6 + 4 = 10
\]

So oxygen balance gives:

\[
2b = 10
\]

Therefore:

\[
b = 5
\]

Step 5: Write the balanced equation.

Substitute \(b = 5\), \(c = 3\), \(d = 4\) into the equation:

\[
\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}
\]

Final Answer:

\[
\boxed{\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}}
\]