Question

A mixture of MgCl2 and NaCl has an initial mass of 0.4015 g. Excess AgNO3(aq) was added and 1.032 g of precipitate, AgCl(s), was recovered. How many moles of MgCl2 were in the original mixture?

Answer 1

Answer:

8.909*10^-4 moles

Explanation:

The mixture contains MgCl$_{2}$ and NaCl with a total mass of 0.4015 g. The mass of precipitate AgCl(aq) is 1.032 g and it has a molar mass of 143.32 g/mol. Therefore, the moles of the precipitate is:

n = 1.032/143.32 = 7.2007*10^-3 moles

Molar mass of NaCl = 58.44 g/mol and the molar mass of MgCl$_{2}$ is 95.21 g/mol. Let the mass (g) of MgCl$_{2}$  in the original mixture be 'x'. Thus:

7.2007*10^-3 = (0.4015-x)/58.44 + 2x/95.21

(7.2007*10^-3)*58.44*95.21 = 95.21(0.4015-x) + 2x(58.44)

40.06505 = 38.227 -95.21x + 116.88x

40.06505 - 38.227 = -95.21x + 116.88x

1.83805 = 21.67x

x = 1.83805/21.67 = 0.0848 g

moles of MgCl$_{2}$ = 0.0848g/95.21 g/mol = 8.909*10^-4 moles