How many grams of CaCl₂ are formed when 30.00 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) A) 0.00395 g B) 0.00789 g C) 0.0158 g D) 0.0253 g

Question

Quizplus · Accepted Answer

First, calculate moles of Ca(OH)₂: 0.00237 M * 0.030 L = 7.11 x 10⁻⁵ moles. From the balanced equation, 1 mole of Ca(OH)₂ produces 1 mole of CaCl₂. Therefore, 7.11 x 10⁻⁵ moles of CaCl₂ are formed. The molar mass of CaCl₂ is approximately 110.98 g/mol. Thus, mass = 7.11 x 10⁻⁵ moles * 110.98 g/mol = 0.00789 g.

How Many Grams of CaCl2 Are Formed When 30

How Many Grams of CaCl₂ Are Formed When 30