Question

Suppose a 12.42g mixture of NaCl and CaCl2 is dissolved in water. To this solution, an excess of AgNO3 solution is added resulting in the formation of 31.12g of the precipitate. Calculate the % by mass of NaCl and CaCl2 in the original mixture. Show all work.

Answer 1

NaCl and CaCl₂ react with AgNO₃ as shown

NaCl(aq) + AgNO3(aq) + AgCl(s) + NaNO3(aq)

CaCl2(aq) + 2AgNO3(aq) + 2AgCl(s) + Ca(NO3)2(aq)

So precipitate formed in both reactions is AgCl

Molar mass of AgCl=Molar mass of Ag+Molar mass of Cl=107.9 g/mol+35.5 g/mol=143.4 g/mol

Number of moles of AgCl formed=Mass/Molar mass

=31.12 g/143.4 g/mol=0.22 mol

Now as per the above equations we can say that ratio of number of moles of AgCl formed from NaCl and that formed from CaCl₂ is 1:2

So assuming p to be number of moles of AgCl formed from NaCl and 2p to be number of moles of AgCl from CaCl₂.

So p+2p=0.22 mol

3p=0.22 mol

p=0.22 mol/3=0.073 mol

So p= 0.073 mol AgCl is formed from NaCl and 2p=2x0.073=0.146 mol AgCl is formed from CaCl₂.

We can see from given balanced chemical equation (1), that 1 mol AgCl is formed from 1 mol NaCl

So 0.073 mol AgCl is formed from 0.073 mol NaCl

Also as per the balanced chemical equation (2),

2 mol AgCl is formed from 1 mol CaCl₂

1 mol AgCl is formed from 1/2 mol CaCl₂

0.146 mol AgCl is formed from (1/2)x0.146 mol=0.073 mol CaCl₂

Molar mass of NaCl=Molar mass of Na+Molar mass of Cl=23 g/mol+35.5 g/mol=58.5 g/mol

So mass of NaCl in the mixture

=number of moles of NaCl x Molar mass of NaCl

=0.073 mol x 58.5 g/mol=4.27 g

Mass percent of NaCl in the mixture

=(Mass of NaCl in the mixture/Mass of mixture)x100

=(4.27 g/12.42 g)x100=34.38%

Mass percent of CaCl₂ in the mixture=100%-34.38%=65.62%