Find the balanced reaction

To find the balanced reaction for the combustion of \( C_3H_3 \), we need to ensure that the number of each type of atom on the reactant side is equal to the number on the product side. The products of the combustion of hydrocarbons are typically carbon dioxide \( CO_2 \) and water \( H_2O \).

Starting with the unbalanced equation:

\[
C_3H_3 + O_2 → CO_2 + H_2O
\]

1. **Count the atoms present in \( C_3H_3 \)**:
- Carbon (C): 3
- Hydrogen (H): 3

2. **Products**:
- Each \( CO_2 \) molecule contains 1 Carbon atom.
- Each \( H_2O \) molecule contains 2 Hydrogen atoms.

3. **Balancing the Carbon**:
- To balance the carbon, we need 3 \( CO_2 \):
\[
C_3H_3 + O_2 → 3CO_2 + H_2O
\]

4. **Balancing the Hydrogen**:
- The 3 Hydrogen atoms in \( C_3H_3 \) require 1.5 \( H_2O \) (because each \( H_2O \) has 2 Hydrogen):
\[
C_3H_3 + O_2 → 3CO_2 + 1.5H_2O
\]

5. **Counting the Oxygens on the product side now**:
- From 3 \( CO_2 \): \( 3 \times 2 = 6 \)
- From 1.5 \( H_2O \): \( 1.5 \times 1 = 1.5 \)
- Total oxygen needed: \( 6 + 1.5 = 7.5 \)

6. **Relating Oxygen in \( O_2 \)**:
- Since \( O_2 \) contains 2 oxygen atoms, we find how many \( O_2 \) molecules are needed:
\[
\frac{7.5}{2} = 3.75 O_2
\]

7. **To eliminate the decimal, we multiply the entire equation by 4:**
\[
4C_3H_3 + 15O_2 → 12CO_2 + 6H_2O
\]

Now let's review the answer choices to find the one that matches what we derived:

a) \( 4C_3H_3 + 15O_2 → 6CO_2 + 3H_2O \) (incorrect)

b) \( 2C_3H_3 + 15O_2 → 6CO_2 + 3H_2O \) (incorrect)

c) \( 4C_3H_3 + 15O_2 → 12CO_2 + 6H_2O \) (correct)

d) \( 4C_3H_3 + 30O_2 → 12CO_2 + 6H_2O \) (incorrect)

The correct balanced reaction is:

**c) \( 4C_3H_3 + 15O_2 → 12CO_2 + 6H_2O \)**