A sample of solid KCl (potassium chloride) weighing 0.500 g is mixed with an unweighed sample of solid MgCl2 (magnesium chloride) and the mixture is then completely dissolved in water to form a clear solution. An aqueous solution of AgNO3 (silver nitrate) is then added to this mixture. A reaction occurs and insoluble solid AgCl (silver chloride) is precipitated. The equation for the reaction is as follows:

Question

A sample of solid KCl (potassium chloride) weighing 0.500 g is mixed with an unweighed sample of solid MgCl2 (magnesium chloride) and the mixture is then completely dissolved in water to form a clear solution. An aqueous solution of AgNO3 (silver nitrate) is then added to this mixture.  A reaction occurs and insoluble solid AgCl (silver chloride) is precipitated.   The equation for the reaction is as follows:

KCl(aq) + MgCl2(aq)+ 3AgNO3(aq) KNO3(aq)+ Mg(NO3)2(aq)+ 3AgCl(s)

When the reaction is over, both of the original chlorides, KCl and MgCl2, are completely used up.  Analysis shows that 8.486 g of AgCl has been produced.  Find the mass of the MgCl2 sample used to make the solid mixture.

Use the following atomic masses: Ag = 107.87,  Cl = 35.45,  K = 39.10,  Mg = 24.31,  N = 14.01

DrBob222 · Answer

I worked this problem for someone last night. Convert 0.500 g KCl to grams AgCl, then subtract from total AgCl obtained (8.486 g). The difference is the AgCl due to MgCl2. Then convert that number (the difference) to grams MgCl2 and you have it.