What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

Options

A. 2.8 kg
B. 6.4 kg
C. 9.6 kg
D. 96 kg

Explanation

To determine the weight of oxygen required for the complete combustion of 2.8 kg of ethylene (C2H4), we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of ethylene.
The complete combustion of ethylene can be represented by the following equation:

\[ \text{C}_2\text{H}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]

Balancing the equation gives:

\[ \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \]

### Step 2: Determine the molar mass of ethylene (C2H4).
The molar mass of ethylene (C2H4) is calculated as follows:

- Carbon (C): 12 g/mol × 2 = 24 g/mol
- Hydrogen (H): 1 g/mol × 4 = 4 g/mol

Total molar mass of C2H4 = 24 g/mol + 4 g/mol = 28 g/mol.

### Step 3: Calculate the moles of ethylene in 2.8 kg.
Convert 2.8 kg of ethylene to grams:

\[ 2.8 \text{ kg} = 2800 \text{ g} \]

Now, calculate the number of moles of ethylene:

\[ \text{Moles of C}_2\text{H}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{2800 \text{ g}}{28 \text{ g/mol}} = 100 \text{ moles} \]

### Step 4: Use the stoichiometry of the balanced equation to find the moles of oxygen required.
From the balanced equation, we see that 1 mole of C2H4 requires 3 moles of O2. Therefore, for 100 moles of C2H4:

\[ \text{Moles of O}_2 = 100 \text{ moles C}_2\text{H}_4 \times 3 = 300 \text{ moles O}_2 \]

### Step 5: Calculate the mass of oxygen required.
The molar mass of O2 is:

- Oxygen (O): 16 g/mol × 2 = 32 g/mol.

Now, calculate the mass of oxygen required:

\[ \text{Mass of O}_2 = \text{moles} \times \text{molar mass} = 300 \text{ moles} \times 32 \text{ g/mol} = 9600 \text{ g} \]

Convert grams to kilograms:

\[ 9600 \text{ g} = 9.6 \text{ kg} \]

### Final Answer:
The weight of oxygen required for the complete combustion of 2.8 kg of ethylene is **9.6 kg**.

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