Question

a 0.5153g solid sample containing only a mixture of LaCl3 & Ce(NO3)3 was dissolved in water. Potassium iodate was then added to the obtained solution producing the precipitates La(IO3)3 molar mass 663.61 g/mol & Ce(IO3)3 molar mass 664.82 g/mol. the total mass of the precipitates was found to be equal to 1.1650g. Calc the % of LaCl3 and Ce(NO3)3 in the sample.

i have to where there are two equations and two unknowns. I just don't know how to solve for the equations

Answer 1

Given,

Mass of the mixture of LaCl3 and Ce(NO3)3 = 0.5153 g

Let there be 'X' g LaCl3

=> Moles of LaCl3 = X / 245.26

Moles of La in 1 mole LaCl3 = 1

=> Moles of La in the mixture = X / 245.26

Mass of Ce(NO3)3 = 0.5153 - X

=> Moles of Ce(NO3)3 = (0.5153 - X) / 326.15

=> Moles of Ce in the mixture = (0.5153 - X) / 326.15

LaCl3 + 3KIO3 -----> La(IO3)3 + 3KCl

and Ce(NO3)3 + 3KIO3 -----> Ce(IO3)3 + 3KNO3

Total Mass of La(IO3)3 and Ce(IO3)3 = 1.165 g

Let Mass of La(IO3)3 = Y

=> Moles of La(IO3)3 = Y / 663.61

=> Moles of La = Y / 663.61

Mass of Ce(IO3)3 = 1.165 - Y

=> Moles of Ce(IO3)3 = (1.165 - Y) / 664.82

=> Moles of Ce = (1.165 - Y) / 664.82

X / 245.26 = Y / 663.61 -------(1)

and (0.5153 - X) / 326.15 = (1.165 - Y) / 664.82 ----(2)

From (1) we get,

Y = 2.71 X

Putting this in (2)

(0.5153 - X) / 326.15 = (1.165 - 2.71 X) / 664.82

=> 2.0383 x (0.5153 - X) = (1.165 - 2.71 X)

=> 1.05 - 2.0383 X = 1.165 - 2.71 X

=> X = 0.1706 g = Mass of LaCl3

=> % LaCl3 = (0.1706 / 0.5153) x 100 = 33.11 %

% Ce(NO3)3 = 100 - 33.11 = 66.89 %