If isobutane and n-butane are present in a gas, then how much oxygen should be required for complete combustion of 5 kg of this gas

Options

A. (a)17.9 kg
B. (b)9 kg
C. (c )27 kg
D. (d)1.8 kg

Explanation

To determine how much oxygen is required for the complete combustion of 5 kg of a gas mixture containing isobutane and n-butane, we can follow these steps:

### Step 1: Understand the Combustion Reaction
The combustion reaction for butane (C4H10) can be represented as:
\[ 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \]
This equation shows that 2 moles of butane react with 13 moles of oxygen.

### Step 2: Calculate the Molar Mass of Butane
The molar mass of butane (C4H10) is calculated as follows:
- Carbon (C): 12 g/mol × 4 = 48 g/mol
- Hydrogen (H): 1 g/mol × 10 = 10 g/mol
- Total molar mass of C4H10 = 48 g/mol + 10 g/mol = 58 g/mol

### Step 3: Determine the Amount of Oxygen Required for One Mole of Butane
From the balanced equation, we see that:
- For 2 moles of C4H10, 13 moles of O2 are required.
- Therefore, for 1 mole of C4H10, the amount of O2 required is:
\[ \frac{13}{2} = 6.5 \text{ moles of } O_2 \]

### Step 4: Calculate the Mass of Oxygen Required for One Mole of Butane
The molar mass of oxygen (O2) is:
- Oxygen (O): 16 g/mol × 2 = 32 g/mol
- Thus, the mass of oxygen required for 1 mole of butane is:
\[ 6.5 \text{ moles} \times 32 \text{ g/mol} = 208 \text{ grams of } O_2 \]

### Step 5: Calculate the Amount of Oxygen Required for 5 kg of Butane
Now, we need to find out how much oxygen is needed for 5 kg (5000 grams) of butane:
- The number of moles of butane in 5 kg is:
\[ \text{Number of moles} = \frac{5000 \text{ g}}{58 \text{ g/mol}} \approx 86.21 \text{ moles of } C4H10 \]

### Step 6: Calculate the Total Oxygen Required
Using the amount of oxygen required for 1 mole of butane, we can find the total oxygen needed:
- Total oxygen required:
\[ 86.21 \text{ moles} \times 6.5 \text{ moles of } O_2 \approx 560.37 \text{ moles of } O_2 \]
- Now, convert moles of O2 to grams:
\[ 560.37 \text{ moles} \times 32 \text{ g/mol} \approx 17973.84 \text{ grams of } O_2 \]
- Converting grams to kilograms:
\[ 17973.84 \text{ g} \approx 17.97 \text{ kg of } O_2 \]

### Conclusion
Thus, the amount of oxygen required for the complete combustion of 5 kg of the gas mixture (isobutane and n-butane) is approximately **17.97 kg**.