A 10g sample of an unknown ionic compound is dissolved, and the solution is treated with enough AgNO3 to precipitate all the chloride ion. If 30.1 g of AgCI are recovered, which of the following compound could be the unknown?

Question

A. NaCI
B. NaNO3
C.BaCI2
D. Mg CI2
E.KCI
Pleas help me to solve this problem, I know the answer but I don't know the step how to solve it.

Answer 1

of 30.1 g of AgCl, how many grams of Cl is that? ANS 30.1*35.4/(35.4+107.8)=7.44 grams. That leaves 2.56g for the metal

Now check moles: molesCl=7.44/35.4=.21moles

Now of the formulas, which metal has .21moles in 2.56g

Try NaCl: .21=2.56/23=not equal
Now, try KCl: .21=2.56/39=wont work.

Try MgCl2. In this, you want .21/2 moles Mg, .105=2.56/24.3=.105
amazing.

Answer 2

lol I know its late but for any other person that looks at this question in the future.

Using molar mass of AgCl you get roughly 24.74% Cl in the ionic compound.

Next, multiply that value in decimal form by 30.1 grams to find the grams of Cl. You should get about 7.44.

Then, you know the percent of Cl in the given 10 grams is .744 so you equate this to the grams of Chloride over grams of Chloride+x. NOTE: for the Magnesium instead of having 35.453 grams you will have 70.906 grams because there are two Cl.

Finally, solve for x and you should get 24.4g, which is fairly close to the molar mass of Magnesium.

Answer 3

good