Stoichiometric decomposition yield: From 90 kg of water, what mass of oxygen (O2) can theoretically be produced? (Use 2 H2O → 2 H2 + O2.)

Introduction / Context:
Simple stoichiometric calculations connect reactant masses to product yields. Water splitting to hydrogen and oxygen is a classic example used to teach mass–mole conversions and reaction stoichiometry constraints.

Given Data / Assumptions:

• Balanced reaction: 2 H2O → 2 H2 + O2.
• Molar masses: H2O = 18 kg/kmol; O2 = 32 kg/kmol.
• Assume 100% conversion and no losses.

Concept / Approach:
The molar stoichiometry shows that 2 moles of water produce 1 mole of oxygen gas. Convert the given water mass to moles, apply stoichiometry to get moles of O2, and convert back to mass.

Step-by-Step Solution:

Verification / Alternative check:
Mass conservation check: 90 kg H2O → 80 kg O2 + 10 kg H2 (since H2 mass = 2.5 kmol * 2 kg/kmol = 5 kg; plus remaining 5 kg? Careful: Hydrogen mass should be 10 kg because n_H2 = 2.5 kmol * 2 kg/kmol = 5 kg—however total product mass equals 80 + 5 = 85 kg; the missing 5 kg indicates we miscounted. Correct: From 5 kmol H2O, H content is 10 kmol H; H2 moles = 5 kmol; m_H2 = 5 kmol * 2 kg/kmol = 10 kg. Then total products 80 + 10 = 90 kg, conserving mass.)

Why Other Options Are Wrong:

• 32 or 64 kg: correspond to using 36 kg or 72 kg of water respectively.
• 90 kg: cannot be all oxygen; hydrogen must also form.
• 72 kg: not consistent with stoichiometry.

Common Pitfalls:
Forgetting to convert between mass and moles; misreading the reaction stoichiometry (2:1 water to oxygen ratio).

Final Answer:
80 kg