How many grams of CaCl₂ are formed when 15.00 mL of 0.00237 mol L^-1 Ca(OH) ₂ reacts with excess Cl₂ gas?
2Ca(OH) ₂(aq) + 2Cl₂(g) → Ca(OCl) ₂(aq) + CaCl₂(s) + 2H₂O(l)

Question

How many grams of CaCl₂ are formed when 15.00 mL of 0.00237 mol L^-1 Ca(OH)₂ reacts with excess Cl₂ gas? 2Ca(OH)₂(aq) + 2Cl₂(g) → Ca(OCl)₂(aq) + CaCl₂(s) + 2H₂O(l)

Quizplus · Accepted Answer

The moles of Ca(OH)₂ are calculated as 15.00 mL * 0.00237 mol/L = 3.555 * 10⁻⁵ mol. According to the balanced equation, 1 mole of Ca(OH)₂ produces 0.5 mole of CaCl₂. Therefore, moles of CaCl₂ = 3.555 * 10⁻⁵ mol * 0.5 = 1.7775 * 10⁻⁵ mol. The molar mass of CaCl₂ is approximately 110.98 g/mol, so the mass of CaCl₂ formed is 1.7775 * 10⁻⁵ mol * 110.98 g/mol = 0.00197 g.

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