Question

How many grams of CaCl2 are formed when 35 mL of .237 M Ca(OH)2 reacts with excess Cl2 gas?

2 Ca(OH)2 (aq) + 2Cl2 (g) yield Ca(OCl)2 (aq) + CaCl2 (s) + 2H20 (l)

Answer 1

lets calculate the mol of Ca(OH)2

volume , V = 35 mL

= 3.5*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.237*3.5*10^-2

= 8.295*10^-3 mol

According to balanced equation

mol of CaCl2 formed = (1/2)* moles of Ca(OH)2

= (1/2)*8.295*10^-3

= 4.148*10^-3 mol

This is number of moles of CaCl2

Molar mass of CaCl2,

MM = 1*MM(Ca) + 2*MM(Cl)

= 1*40.08 + 2*35.45

= 110.98 g/mol

use:

mass of CaCl2,

m = number of mol * molar mass

= 4.148*10^-3 mol * 1.11*10^2 g/mol

= 0.4603 g

Answer: 0.460 g