Part A How many grams of CaCl2 are formed when 1500 mL of 00

Part A 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 H20(1) O 0.00197 g O 0.00789 g O 0.0507 g O 0.00394 9 Submit Provide Feedback

Solution

Molarity = moles of solute / liters of solution

therefore moles of Ca(OH)₂ present for reaction are = .00237*15*10^-3

As Cl₂ is in excess so all available  Ca(OH)₂ will react to give  CaCl₂  

so by stoichiometry of the reaction 2 moles of  Ca(OH)₂ give 1 mole of CaCl₂ so available moles will give =

1/2*( .00237*15*10^-3) moles of CaCl₂

Grams of CaCl₂ = moles*molecular weight of CaCl₂

molecular weight of CaCl₂ = 110

therefore Grams of CaCl₂ = 0.00197 gm