Question

A 1.249g mixture containing calcium chloride & sodium chloride reacts w/ excess? sodium carbonate as per equation ..(i) to precipitate 0.567g of calcium carbonate a. Calculate the moles of calcium carbonate precipitated b. How many moles of calcium chloride are present in the mixture c. covert moles of calcium chloride to grams of calcium chloride d. calculate % calcium chloride in th

Answer 1

CaCl2 + Na2CO3   -----------------> CaCO3 + 2 NaCl

1 mol Na2CO3   gives   --------- 1 mol CaCO3

mass = 0.567 g

molar mass of CaCO3 = 100 g/mol

a)

moles of CaCO3 = mass / molar mass = 0.567 / 100 = 5.67 x 10^-3

b)

moles of CaCl2 = 5.67 x 10^-3

c)

mass of CaCl2 = 5.67 x 10^-3 x 110.98

                        = 0.6293 g

d)

% CaCl2 = (mass of CaCl2 / mass of mixture ) x 100

               = (0.6293 / 1.249 ) x 100

               = 50.38%