What is the mass of oxygen that is required for the complete combustion of 2.8 kg ethylene :-

Options

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

Explanation

To find the mass 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 balanced equation for the combustion of ethylene is:
\[ \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \]

### Step 2: Determine the molar mass of ethylene (C2H4).
The molar mass of ethylene can be calculated as follows:
- Carbon (C) has a molar mass of approximately 12 g/mol.
- Hydrogen (H) has a molar mass of approximately 1 g/mol.

Calculating the molar mass:
\[ \text{Molar mass of C}_2\text{H}_4 = (2 \times 12) + (4 \times 1) = 24 + 4 = 28 \text{ g/mol} \]

### Step 3: Determine the molar mass of oxygen (O2).
The molar mass of oxygen (O2) is:
\[ \text{Molar mass of O}_2 = 2 \times 16 = 32 \text{ g/mol} \]

### Step 4: Find the stoichiometric ratio of ethylene to oxygen.
From the balanced equation, 1 mole of ethylene reacts with 3 moles of oxygen. Therefore, the ratio is:
\[ \text{C}_2\text{H}_4 : \text{O}_2 = 1 : 3 \]

### Step 5: Calculate the amount of oxygen needed for 1 mole of ethylene.
Since 1 mole of C2H4 (28 g) requires 3 moles of O2 (96 g):
\[ 1 \text{ mole of C}_2\text{H}_4 \text{ (28 g)} \text{ requires } 3 \text{ moles of O}_2 \text{ (96 g)} \]

### Step 6: Calculate the mass of oxygen required for 2.8 kg of ethylene.
First, convert 2.8 kg of ethylene to grams:
\[ 2.8 \text{ kg} = 2800 \text{ g} \]

Now, find how many moles of ethylene are in 2800 g:
\[ \text{Moles of C}_2\text{H}_4 = \frac{2800 \text{ g}}{28 \text{ g/mol}} = 100 \text{ moles} \]

Using the stoichiometric ratio, calculate the moles of O2 required:
\[ \text{Moles of O}_2 = 100 \text{ moles of C}_2\text{H}_4 \times 3 = 300 \text{ moles of O}_2 \]

Now, convert moles of O2 to grams:
\[ \text{Mass of O}_2 = 300 \text{ moles} \times 32 \text{ g/mol} = 9600 \text{ g} \]

### Step 7: Convert grams of oxygen to kilograms.
\[ 9600 \text{ g} = 9.6 \text{ kg} \]

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

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