How many grams of CaCl₂ are formed when 35.00 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

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 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:35.00 mL x 0.00237 M = 0.00008345 molTherefore, the number of moles of CaCl₂ produced is also 0.00008345 mol.The molar mass of CaCl₂ is 110.98 g/mol.Therefore, the mass of CaCl₂ produced is:0.00008345 mol x 110.98 g/mol = 0.00921 gTherefore, the correct answer is B) 0.00921 g.

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How Many Grams of CaCl₂ Are Formed When 35