Suppose that 0.205 g of a mixture containing Na2SO4 and K2SO4 is dissolved in water. An excess

Question

of BaCl2 is added and a precipitate of 0.298 g of BaSO4 is produced. Calculate the percent by mass
of Na2SO4 and K2SO4 in the original mixture.

Answer 1

This is a two equation problem and the two are to be solved simultaneously. These are difficult to explain by writing so I'll do it in steps.
Let x = mass Na2SO4
and y = mass K2SO4
---------------------
equation 1 is x + y = 0.205

The second equation comes from the ppt of BaSO4. The idea is that you set up the mass BaSO4 from x (the Na2SO4) and the mass BaSO4 from y (the K2SO4), add them together and the sum will be 0.298g BaSO4. To avoid typing and the space used up let MM = molar mass and AM = atomic mass.
amount from Na2SO4 is x*(MM BaSO4/MM Na2SO4)
amount from K2SO4 is y(MM BaSO4/MM K2SO4)
Therefore, the second equation is
x*(MM BaSO4/MM Na2SO4) + y(MM BaSO4/MM K2SO4) = 0.298

Solve those two equations for x and y, then
%Na2SO4 = (x/mass sample)*100 = ?
%K2SO4 = (y/mass sample)*100 = ?

Post your work if you stuck.