A mixture consisting of only iron(II) chloride (FeCl2) and chromium(II) chloride (CrCl2) weighs 1.0294 g. When the mixture is dissolved in water and an excess of silver nitrate is added, all the chloride ions associated with the original mixture are precipitated as insoluble silver chloride (AgCl). The mass of the silver chloride is found to be 2.3609 g. Calculate the mass percentages of iron(II) chloride and chromium(II) chloride in the original mixture. Mass percent FeCl2 = % Mass percent CrCl2 = %

Mass FeCl2 = 0.0333g

Mass CrCl2 = 0.9961g

Explanation:

To solve this problem. The first equation we can write is:

Mass FeCl2 + Mass CrCl2 = 1.0294g (1)

Now, the Chlorides of FeCl2 and CrCl2 react producing 2.3609g of AgCl

Using molar mass of these species (126.75g/mol, 122.9g/mol, 143.32g/mol, respectively), you can write the equation:

2Mass FeCl2 / 126.75 + 2Mass CrCl2 / 122.9 = 2.3609/143.32

That is: Moles Chloride before = Moles Chloride in AgCl after reaction

7.8895x10⁻³Mass FeCl2 + 0.0162734MassCrCl2 = 0.016473 (2)

Replacing (1) in (2):

7.8895x10⁻³ (1.0294g - MassCrCl2) + 0.0162734MassCrCl2 = 0.016473

8.12145x10⁻³ -7.8895x10⁻³MassCrCl2 + 0.0162734Mass CrCl2 = 0.016473

8.3839x10⁻³ MassCrCl2 = 8.35155x10⁻³

Mass CrCl2 = 0.9961g

And:

Mass FeCl2 = 1.0294g - 0.9961g

Mass FeCl2 = 0.0333g

idk i’m struggling too