Given:
Mass of ethylene = 2.8 kg
To find:
Weight of oxygen required for complete combustion of ethylene

Solution:
The combustion of ethylene (C2H4) can be represented by the following balanced chemical equation:

C2H4 + O2 → CO2 + H2O

From the equation, we can see that one molecule of ethylene reacts with one molecule of oxygen to produce one molecule of carbon dioxide and one molecule of water.

Step 1: Calculate the molar mass of ethylene (C2H4)
The molar mass of carbon (C) = 12.01 g/mol
The molar mass of hydrogen (H) = 1.008 g/mol

Molar mass of ethylene = (2 * molar mass of carbon) + (4 * molar mass of hydrogen)
= (2 * 12.01 g/mol) + (4 * 1.008 g/mol)
= 24.02 g/mol + 4.032 g/mol
= 28.052 g/mol

Step 2: Convert the mass of ethylene to moles
Moles of ethylene = Mass of ethylene / Molar mass of ethylene
= 2.8 kg / 28.052 g/mol
= 100 mol

Step 3: Determine the stoichiometry between ethylene and oxygen
From the balanced equation, we can see that one molecule of ethylene reacts with one molecule of oxygen.

Step 4: Calculate the molar mass of oxygen (O2)
The molar mass of oxygen (O2) = 2 * molar mass of oxygen
= 2 * 16.00 g/mol
= 32.00 g/mol

Step 5: Calculate the weight of oxygen required
Weight of oxygen required = Moles of ethylene * Molar mass of oxygen
= 100 mol * 32.00 g/mol
= 3200 g
= 3.2 kg

Therefore, the weight of oxygen required for complete combustion of 2.8 kg of ethylene is 3.2 kg.