AP Chem

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

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asked by benji
  1. Wouldn't this work. Check my thinking.
    Calculate how much AgCl can be pptd from 0.500 g KCl. That will be
    0.500 g KCl x (molar mass AgCl/molar mass KCl) = ?? g AgCl.
    Subtract 8.486 - ??g AgCl from the above calcn to arrive at the amount AgCl that must be due to MgCl2 alone.
    Then convert g AgCl that must be due to MgCl2 to g MgCl2 and you have it solved. I did a very quick quickie on the numbers and came up with approximately 2.7 grams MgCl2 but you need to check that carefully.

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    posted by DrBob222
  2. I ended up getting 1.24 g?

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    posted by benji
  3. My answer, after reviewing it, is 2.499 grams MgCl2.
    I used 95.21 for molar mass MgCl2.
    143.32 for molar mass AgCl
    74.55 for molar mass KCl.

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    posted by DrBob222
  4. i keep getting 1.24 g i first coverted 0.500g KCl to g of AgCl g and i got 2.88 g. then i subtracted that from 8.486 and i got 5.602 AgCl. then i converted 5.602 g of AgCl to g of MgCl2 and i got 1.24 g.
    are your numbers similar to that?

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    posted by benji

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