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

Question

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

Quizplus · Accepted Answer

The moles of Ca(OH)₂ are calculated as (0.00237 mol/L) * (0.015 L) = 3.555 x 10⁻⁵ mol. From the balanced equation, 1 mole of Ca(OH)₂ produces 0.5 moles of CaCl₂. Therefore, moles of CaCl₂ = 3.555 x 10⁻⁵ mol * 0.5 = 1.7775 x 10⁻⁵ mol. The molar mass of CaCl₂ is approximately 110.98 g/mol, so the mass of CaCl₂ formed is 1.7775 x 10⁻⁵ mol * 110.98 g/mol = 0.00197 g.

How Many Grams of CaCl2 Are Formed When 15

How Many Grams of CaCl₂ Are Formed When 15