What volume of `O_(2)` at 2.00 atm pressure and `27^(@)C` is requried to burn 10.0 g of heptane `(C_(7)H_(16))` ?
`C_(7)H_(16) + 11 O_(2) to 7 CO_(2) + 8 H_(2)O`

Explanation

To solve the problem of determining the volume of \( O_2 \) required to burn 10.0 g of heptane (\( C_7H_{16} \)), we will follow these steps:

### Step 1: Calculate the number of moles of heptane
First, we need to find the molar mass of heptane (\( C_7H_{16} \)):
- Carbon (C) has a molar mass of approximately 12.01 g/mol.
- Hydrogen (H) has a molar mass of approximately 1.008 g/mol.

Calculating the molar mass of heptane:
\[ \text{Molar mass of } C_7H_{16} = (7 \times 12.01) + (16 \times 1.008) = 84.16 \text{ g/mol} \]

Now, we can calculate the number of moles of heptane in 10.0 g:
\[ \text{Number of moles of } C_7H_{16} = \frac{\text{mass}}{\text{molar mass}} = \frac{10.0 \text{ g}}{84.16 \text{ g/mol}} \approx 0.119 \text{ moles} \]

### Step 2: Use the stoichiometry of the reaction
From the balanced chemical equation:
\[ C_7H_{16} + 11 O_2 \rightarrow 7 CO_2 + 8 H_2O \]
We see that 1 mole of heptane requires 11 moles of \( O_2 \). Therefore, the moles of \( O_2 \) required to burn 0.119 moles of heptane is:
\[ \text{Moles of } O_2 = 0.119 \text{ moles of } C_7H_{16} \times 11 \text{ moles of } O_2 = 1.309 \text{ moles of } O_2 \]

### Step 3: Use the Ideal Gas Law to find the volume of \( O_2 \)
The Ideal Gas Law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

Convert the temperature from Celsius to Kelvin:
\[ T = 27^\circ C + 273.15 = 300.15 \text{ K} \]

Now, substituting the known values into the Ideal Gas Law:
\[ V = \frac{nRT}{P} = \frac{(1.309 \text{ moles}) \times (0.0821 \text{ L·atm/(K·mol)}) \times (300.15 \text{ K})}{2.00 \text{ atm}} \]

Calculating the volume:
\[ V = \frac{(1.309) \times (0.0821) \times (300.15)}{2.00} \approx \frac{32.1}{2.00} \approx 16.05 \text{ L} \]

### Final Answer
The volume of \( O_2 \) required to burn 10.0 g of heptane at 2.00 atm pressure and 27°C is approximately **16.05 liters**.

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