CaCl2 + K2CO3 --> CaCO3 + 2KCl moles CaCl2 = 2.0 g/110.978 g/mol = 0.018 mol mol

ID: 699889 • Letter: C

Question

CaCl2 + K2CO3 --> CaCO3 + 2KCl

moles CaCl2 = 2.0 g/110.978 g/mol = 0.018 mol

moles K2CO3 = 2.5 g/138.205 g/mol = 0.0181 mol

since moles of CaCl2 is lower, this is the limiting reactant

Theoretical yield of CaCO3 = 0.018 mol x 100.086 g/mol = 1.80155 g

Actual yield of CaCl2 = 1.8 g

Percent yield = 1.80 x 100/1.80155 = 99.914 %

moles Ca2+ in initial solution based on actual yield = 0.9914 x 0.018 mol = 0.017845 mol

mass CaCl2 in initial solution based on actual yield = 110.978 g/mol x 0.017845 mol = 1.980 g

concentration of CaCl2 in %w/v = 1.980 g x 100/volume of solution (not given above)

mass of water associated with CaCl2 = 3.5 - 2.0 = 1.5 g

CaCl2: 2.0g

K2Co3: 2.5g

Filter paper: 2.2g

Watch Glass: 33.2g

Precipitate: 1.8g

24 Hour CaCl2: 3.5g

Calculations needed

Theoretical yield (CaCO3):

Actual yield (CaCO3):

Percent yield:

Moles of Ca present in original solution, based on actual yield:

Mass of CaCl2 present in original solution, based on actual yield:

Analyze the data and determine the actual concentration of calcium chloride in the solution. Show all calculations

and report in % wt/v concentration.

Determine the mass of water which became associated with the calcium chloride during its exposure to

the air.