When pentane , `C_(5) H_(12)` is burned in excess oxygen , the products of the reaction are `CO_(2)(g)` and `H_(2)O (l)` . The balanced equation for this combustion is `C_(5) H_(12)(g) + xO_(2)(g) rarr 5CO_(2)(g) + 6 H_(2) O (l)`
The coefficient (x) of oxygen should be

Options

A. 16
B. 12
C. 11
D. 8

Answer

To find the coefficient (x) of oxygen in the combustion reaction of pentane (C₅H₁₂), we will follow these steps:

### Step 1: Write the unbalanced combustion equation
The combustion of pentane can be represented as:
\[ C_5H_{12} + x O_2 \rightarrow 5 CO_2 + 6 H_2O \]

### Step 2: Count the number of carbon atoms
In pentane (C₅H₁₂), there are 5 carbon atoms. Therefore, in the products, we will have 5 CO₂ molecules, which also contain 5 carbon atoms. This part of the equation is balanced.

### Step 3: Count the number of hydrogen atoms
Pentane has 12 hydrogen atoms. In the products, each water molecule (H₂O) contains 2 hydrogen atoms. Therefore, 6 water molecules will contain:
\[ 6 \times 2 = 12 \text{ hydrogen atoms} \]
Thus, the hydrogen atoms are also balanced.

### Step 4: Count the number of oxygen atoms in the products
Now, we need to count the total number of oxygen atoms produced:
- From 5 CO₂: \( 5 \times 2 = 10 \) oxygen atoms
- From 6 H₂O: \( 6 \times 1 = 6 \) oxygen atoms

Total oxygen atoms in the products:
\[ 10 + 6 = 16 \text{ oxygen atoms} \]

### Step 5: Relate the oxygen atoms to the reactants
In the reactants, each O₂ molecule contains 2 oxygen atoms. Therefore, to find the coefficient (x) of O₂, we set up the equation:
\[ 2x = 16 \]

### Step 6: Solve for x
To find x, we divide both sides by 2:
\[ x = \frac{16}{2} = 8 \]

### Final balanced equation
The balanced equation for the combustion of pentane is:
\[ C_5H_{12} + 8 O_2 \rightarrow 5 CO_2 + 6 H_2O \]

Thus, the coefficient (x) of oxygen is **8**.

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