Stoichiometric combustion of ethylene (C2H4) – products per kilogram Consider the complete (stoichiometric) combustion of 1 kg of ethylene, C2H4. It produces 22/7 kg of carbon dioxide. What is the corresponding mass of water (liquid or steam) formed?

Introduction / Context:
Stoichiometric combustion problems convert a fuel's chemical formula into precise product and reactant masses. Ethylene (C2H4) is a common unsaturated hydrocarbon; balancing its combustion gives fixed product ratios independent of air excess.

Given Data / Assumptions:

• Chemical formula: C2H4.
• Complete combustion to CO2 and H2O.
• Basis: 1 kg of fuel; standard atomic masses C = 12, H = 1, O = 16.

Concept / Approach:

Balance the reaction on a molar basis, then convert to mass per kilogram of fuel. Combustion equation: C2H4 + 3 O2 → 2 CO2 + 2 H2O. From this, two moles of CO2 and two moles of H2O are formed per mole of C2H4 burned. Convert moles to masses and scale by the molar mass of ethylene.

Step-by-Step Solution:

Verification / Alternative check:

The given CO2 yield 22/7 kg per kg fuel matches the calculation, confirming consistent stoichiometry. Therefore, the water formed must be 9/7 kg per kg fuel.

Why Other Options Are Wrong:

11/7, 7/4, 11/4, 5/7 do not satisfy the mass ratios implied by the balanced equation; only 9/7 aligns with 2 moles of water per mole of ethylene.

Common Pitfalls:

Using oxygen requirement data to infer products directly; products are fixed by fuel composition under complete combustion, independent of excess air.

Final Answer:

9/7