[JessicaSource]

Hey guys, this is one of the questions from my chemistry book assignments.  It is part of my suggested problems for my upcoming test next week.   Any help is appreciated.

A solution contains Cr(+3) ion and Mg(+2) ion.  The addition of 1.00L of 1.51 M NaF solution is required to cause the complete precipitation of these ions as CrF3(s) and MgF2(s). The total mass of the precipitate is 49.9g

Find the mass of the Cr(+3) in the original solution as grams

I know that there are two ions here and 1.00 Liter of NaF is being added to it. Since Molarity is moles per liter, I formulated that 1.51 moles were being added to it, or 63.4049 grams NaF.

I get that 49.9 grams is the precipitate (which I know in this context means the solid) but doesn't that include CrF3 AND MgF2? (As they are both solids?)

How do I get the grams of Cr(+3)?

I thank you for your help in this question for me. I also thank you for this nice forum.  It is very informative.

[Astrokel]

hello,

first by writting off the balanced ionic equation of the two reactions. next, you have to form one equation by utilizing the moles of F^- added and also mass of NaF. Hint: Let x be the moles of F^- used in one of the reaction, so the amount of F^- used in the other reaction must be (1.51-x).

[Borek]

I get that 49.9 grams is the precipitate (which I know in this context means the solid) but doesn't that include CrF3 AND MgF2? (As they are both solids?)

Yes, this is sum of their masses.

Assume x moles of one and y moles of the second ion. Write expressions for the mass of precipitate and for the number of moles of precipitate. That yields two equations with two unknowns. Solve.

Note: following Astrokel hints you will end doing the same, just in slightly different way.