Lime chemistry — theoretical water for slaking quicklime For slaking 10 kg of CaO (quicklime), what is the theoretical mass of water required to form Ca(OH)2?

Introduction / Context:
Slaking of quicklime is a fundamental reaction in building materials and environmental engineering. Correct water dosage is important for efficient hydration, heat control, and achieving desired workability in lime mortars and stabilization works.

Given Data / Assumptions:

• Chemical reaction: CaO + H2O → Ca(OH)2.
• Molar masses: CaO = 56 kg/kmol, H2O = 18 kg/kmol.
• Target quicklime mass: 10 kg.

Concept / Approach:
The stoichiometric water requirement follows directly from molar mass ratios in the balanced equation. One mole of water reacts with one mole of CaO. Compute water mass per mass of CaO and scale to 10 kg.

Step-by-Step Solution:
Required water per kg CaO = 18 / 56 kg/kg.For 10 kg CaO: water = (18 / 56) * 10.Calculate: 18 / 56 = 0.321428…Water needed = 0.321428… * 10 = 3.21428… kg ≈ 3.2 kg.

Verification / Alternative check:
Cross-check by proportionality: 56 kg CaO requires 18 kg H2O; 10 kg CaO requires (10 * 18 / 56) = 3.214 kg.

Why Other Options Are Wrong:
2.2 kg and 1.5 kg understate the stoichiometric need and would leave unreacted CaO.

4.0 kg exceeds the theoretical requirement; extra water may be used in practice for workability but is not the theoretical stoichiometric amount.

“None of these” is incorrect because 3.2 kg is available and correct.

Common Pitfalls:
Confusing theoretical water with practical water addition; ignoring heat of hydration which can cause flash setting and safety hazards; not accounting for lime purity in real projects.

Final Answer:
3.2 kg