Butane gas burns in air according to the following reaction.
`2C_(4)H_(10) +13O_(2) rarr 10 H_(2)O +8 CO_(2)`.
Suppose the initial and final temperature are equal and high enough so that all reactants and products act as perfect gases. Two moles of butane ar mixed with `13` moles of oxygen and then completely reacted. Find the final pressure (if the volume remains unchanged and the pressure before reaction is `P_(0))`?

Answer

To solve the problem, we will follow these steps:

### Step 1: Write down the balanced chemical equation
The balanced chemical reaction for the combustion of butane is:
\[ 2C_{4}H_{10} + 13O_{2} \rightarrow 10H_{2}O + 8CO_{2} \]

### Step 2: Identify the initial and final moles of gases
- **Initial moles:**
- Moles of butane (\(C_{4}H_{10}\)): 2 moles
- Moles of oxygen (\(O_{2}\)): 13 moles
- Total initial moles (\(N_{initial}\)): \(2 + 13 = 15\) moles

- **Final moles:**
- Moles of water (\(H_{2}O\)): 10 moles
- Moles of carbon dioxide (\(CO_{2}\)): 8 moles
- Total final moles (\(N_{final}\)): \(10 + 8 = 18\) moles

### Step 3: Use the ideal gas law
According to the ideal gas law, we have:
\[ PV = nRT \]
Since the volume \(V\) and temperature \(T\) are constant, we can relate the pressures and moles before and after the reaction:
\[ \frac{P_{initial}}{N_{initial}} = \frac{P_{final}}{N_{final}} \]

### Step 4: Substitute the known values
We know:
- \(P_{initial} = P_{0}\)
- \(N_{initial} = 15\)
- \(N_{final} = 18\)

Substituting these values into the equation gives:
\[ \frac{P_{0}}{15} = \frac{P_{final}}{18} \]

### Step 5: Solve for \(P_{final}\)
Cross-multiplying gives:
\[ P_{final} = P_{0} \cdot \frac{18}{15} \]
\[ P_{final} = P_{0} \cdot \frac{6}{5} \]

### Final Answer
Thus, the final pressure \(P_{final}\) is:
\[ P_{final} = \frac{6}{5} P_{0} \]

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