Question

a 12.42g sample of a mixture of NaCl and CaCl was dissolved in water. Excess AgNO3 was added and all of the chloride was precipitated as AgCl. 31.70g of AgCl was collected. Calculate the w/w% of NaCl in the orginal mixture.

Answer 1

mol of AgCl = mass/MW = 31.70/143.32 = 0.22118

mol of Cl- = 0.22118

then

mol of NaCl + 2*CaCl2 = 0.22118

mass balance:

mass of NaCl + mass of CaCl2 = 12.42

change to mol

let "X" be the mass of NaCl so

mass of NaCl = mol of NaCl/MW o f NACl = mass of NaCl / 58.5 = x/58.5

mass of CaCl2 = mol of CaCl2 /MW o f CaCl2 = mass of CaCl2 / 110.98 = (12.42-x) / 110.98

mol of NaCl + 2*CaCl2 = 0.22118

x/58.5 + 2*(12.42-x) / 110.98 = 0.22118

get x

x*110.98 + 58.5 *2*(12.42-x) = 0.22118*110.98 *58.5  

110.98x+ 1453.14-117x = 1435.97

x(110.98-117) = 1435.97-1453.14

x = (1435.97-1453.14 )/ (110.98-117)

x = 2.8521 g of NaCl

% w/w = 2.8521/(12.42) * 100 = 22.963% of NaCL