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Balance the entire chemical reaction using an atom inventory. Remember to balance each element in the reaction.

[tex]\[ C_3H_8 + O_2 \rightarrow CO_2 + H_2O \][/tex]

When the reaction is balanced, what is the correct whole number coefficient for [tex]\[ O_2 \][/tex]?

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GinnyAnsweridn
To balance the chemical equation [tex]\( C_3H_8 + O_2 \rightarrow CO_2 + H_2O \)[/tex], we need to ensure that the number of atoms of each element on the reactant side equals the number of atoms on the product side. We will follow these steps:

### Step 1: Write Down the Unbalanced Equation
[tex]\[ C_3H_8 + O_2 \rightarrow CO_2 + H_2O \][/tex]

### Step 2: Count the Number of Atoms of Each Element
- On the reactant side:
- Carbon (C): 3 (from [tex]\( C_3H_8 \)[/tex])
- Hydrogen (H): 8 (from [tex]\( C_3H_8 \)[/tex])
- Oxygen (O): 2 (from [tex]\( O_2 \)[/tex])

- On the product side:
- Carbon (C): 1 (from [tex]\( CO_2 \)[/tex])
- Hydrogen (H): 2 (from [tex]\( H_2O \)[/tex])
- Oxygen (O): 3 (2 from [tex]\( CO_2 \)[/tex] and 1 from [tex]\( H_2O \)[/tex])

### Step 3: Balance Carbon Atoms
We have 3 carbon atoms on the reactant side and only 1 on the product side. Hence, we need to place a coefficient of 3 in front of [tex]\( CO_2 \)[/tex] to balance the carbon atoms:
[tex]\[ C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O \][/tex]

### Step 4: Balance Hydrogen Atoms
We have 8 hydrogen atoms on the reactant side and only 2 on the product side. Hence, we need to place a coefficient of 4 in front of [tex]\( H_2O \)[/tex] to balance the hydrogen atoms:
[tex]\[ C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O \][/tex]

### Step 5: Balance Oxygen Atoms
Now, we need to balance the oxygen atoms. We now have:
- On the reactant side: 2 oxygen atoms (from [tex]\( O_2 \)[/tex])
- On the product side: [tex]\( 3 \times 2 + 4 \times 1 = 6 + 4 = 10 \)[/tex] oxygen atoms

To balance the oxygen atoms, we need 10 oxygen atoms on the reactant side. Since each [tex]\( O_2 \)[/tex] molecule provides 2 oxygen atoms, we need [tex]\(\frac{10}{2} = 5 \)[/tex] [tex]\( O_2 \)[/tex] molecules:
[tex]\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \][/tex]

Now, the equation is balanced:
- Carbon: 3 on both sides
- Hydrogen: 8 on both sides
- Oxygen: 10 on both sides

### Conclusion
The whole number coefficient for [tex]\( O_2 \)[/tex] when the chemical equation is balanced is [tex]\( \boxed{5} \)[/tex].
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