How many grams of CaCl₂ are formed when 25.00 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)0.00329 g B)0.00658 g C)0.0132 g D)0.0304 g

Question

Quizplus · Accepted Answer

First, calculate the moles of Ca(OH)₂: 0.00237 M * 0.025 L = 5.925 x 10⁻⁵ moles. According to the balanced equation, 1 mole of Ca(OH)₂ produces 1 mole of CaCl₂. Therefore, 5.925 x 10⁻⁵ moles of CaCl₂ are formed. The molar mass of CaCl₂ is approximately 110.98 g/mol. Thus, the mass of CaCl₂ formed is 5.925 x 10⁻⁵ moles * 110.98 g/mol = 0.00657 g, which rounds to 0.00658 g.

How Many Grams of CaCl2 Are Formed When 25

How Many Grams of CaCl₂ Are Formed When 25