You dissolve 150.0 grams of a mixture of MgCl₂ and KCl in water, add a solution of excess AgNO₃, and precipitate all of the chloride ion as AgCl. After filtration and drying, you find that you have formed 329.0 grams of AgCl. What is the mass in grams of KCl that was in the original 150.0 grams of mixture?

112.6g KCl

Explanation:

With the mass of AgCl that we obtain after the precipitation we can obtain its moles = Moles Cl. As all Cl comes from KCl, moles of Cl = Moles KCl + 2*Moles MgCl₂. We can write:

Moles AgCl = Moles Cl

329.0g * (1mol / 143.32g) =

2.2956 moles Cl = Moles KCl + 2 Moles MgCl₂ (1)

150.0g = MolesKCl * 74.5513g/mol + Moles MgCl₂*95.211g/mol (2)

Replacing (2) in (1):

2.2956 moles Cl = 150/74.5513 - 95.211/74.5513 Moles MgCl₂ + 2 Moles MgCl₂

2.2956 = 2.012 - 1.2771Moles MgCl₂ + 2moles MgCl₂

0.2836 = 0.7229 moles MgCl₂

0.392= moles MgCl₂

Mass MgCl₂:

0.392 * (95.211g/mol) = 37.4g MgCl₂

Mass KCl:

KCl = 150.0g - 37.4g =

the answer is barium and copper bromide