How many grams of CaCl₂ are formed when 15.00 mL of 0.00237 M Ca(OH) ₂ 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 M Ca(OH)₂ 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 balanced chemical equation is:2 Ca(OH)₂(aq)+ 2 Cl₂(g)→ Ca(OCl)₂(aq)+ CaCl₂(s)+ 2 H₂O(l)From the balanced chemical equation, we can see that 2 moles of Ca(OH)₂ react with 2 moles of Cl₂ to produce 1 mole of CaCl₂.Therefore, the number of moles of CaCl₂ produced is equal to the number of moles of Ca(OH)₂ used.The number of moles of Ca(OH)₂ used is:n(Ca(OH)₂) = V(Ca(OH)₂) × C(Ca(OH)₂) = 15.00 mL × 0.00237 M = 0.00003555 molTherefore, the number of moles of CaCl₂ produced is:n(CaCl₂) = n(Ca(OH)₂) = 0.00003555 molThe molar mass of CaCl₂ is 110.98 g/mol.Therefore, the mass of CaCl₂ produced is:m(CaCl₂) = n(CaCl₂) × M(CaCl₂) = 0.00003555 mol × 110.98 g/mol = 0.00394 gTherefore, the correct answer is A) 0.00394 g.

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How Many Grams of CaCl₂ Are Formed When 15