Chem Hwk Questions Theoretical Yield ?

1.
Molar mass of KCl = 39.10 + 35.45 = 74.55 g/mol
Molar mass of O₂ = 16.00×2 = 32.00 g/mol
Molar mass of KClO₃ = 39.10 + 35.45 + 16.00×3 = 122.55 g/mol

Equation for the reaction:
2KCl + 3O₂ → 2KClO₃

If KCl completely reacts:
(100.0 g KCl) × (1 mol KCl / 74.55 g KCl) × (2 mol KClO₃ / 2 mol KCl) × (122.55 g ClO₃ / 1 mol KClO₃)
= 164.4 g KClO₃

If O₂ completely reacts:
(100.0 g O₂) × (1 mol O₂ / 32.00 g O₂) × (2 mol KClO₃ / 3 mol O₂) × (122.55 g KClO₃ / 1 mol KClO₃)
= 255.3 g KClO₃

The calculation base on the complete reaction of KCl gives a smaller amount of product.
Hence, the limiting reactant is KCl, and the theoretical yield is 164.4 g.

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2.
Molar mass of Ca(OH)₂ = 40.08 + 16.00×2 + 1.01×2 = 74.10 g/mol
Molar mass of H₃PO₄ = 1.01×3 + 30.97 + 16.00×4 = 98.00 g/mol
Molar mass of Ca₃(PO₄)₂ = 40.08×3 + 30.97×2 + 16.00×8 = 310.18 g/mol

Equation for the reaction:
3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂O

If Ca(OH)₂ completely reacts:
(10.00 g Ca(OH)₂) × (1 mol Ca(OH)₂ / 74.10 g Ca(OH)₂) × (1 mol Ca₃(PO₄)₂ / 3 mol Ca(OH)₂) × (310.18 g Ca₃(PO₄)₂ / 1 mol Ca₃(PO₄)₂)
= 13.95 g Ca₃(PO₄)₂

If H₃PO₄ completely reacts:
(10.00 g H₃PO₄) × (1 mol H₃PO₄ / 98.00 g H₃PO₄) × (1 mol Ca₃(PO₄)₂ / 2 mol H₃PO₄) × (310.18 g Ca₃(PO₄)₂ / 1 mol Ca₃(PO₄)₂)
= 15.83 g Ca₃(PO₄)₂


The calculation base on the complete reaction of Ca(OH)₂ gives a smaller amount of product.
Hence, the limiting reactant is Ca(OH)₂, and the theoretical yield of Ca₃(PO₄)₂ is 13.95 g.