Question

A 0.5222 g solid sample containing a mixture of LaCl3 (molar mass = 245.26 g/mol) and Ce(NO3)3 (molar mass = 326.13 g/mol) was dissolved in water. The solution was titrated with KIO3 , producing the precipitates La(IO3)3 and Ce(IO3)3 . For the complete titration of both La3+ and Ce3+ , 42.81 mL of 0.1250 M KIO3 was required. Calculate the mass fraction of La and Ce in the sample.

Answer 1

Answer:

Mole fraction of La is 19,58 wt% and of Ce is 28,11 wt%

Explanation:

The reactions of LaCl₃ and Ce(NO₃)₃ with KIO₃ are:

LaCl₃ + 3KIO₃ → La(IO₃)₃ + 3KCl

Ce(NO₃)₃ + 3KIO₃ → Ce(IO₃)₃ + 3KNO₃

The moles required for a complete titration of the sample are:

0,04281L× 0,1250M = 5,351x10⁻³ moles of KIO₃

By the reactions, and defining X are grams of LaCl₃ and Y as grams of Ce(NO₃)₃, these moles are:

5,351x10⁻³ = $\frac{3X}{245,26g/mol}$ + $\frac{3Y}{326,13g/mol}$ (1)

As total mass of the solid sample are 0,5222g:

0,5222g = X + Y (2)

Replacing (2) in (1):

Y = 0,3417g

Thus:

X = 0,1805g

Grams of La are:

0,1805gLaCl₃×$\frac{1mol}{245,26g}$×$\frac{138,90547g}{1molLa}$ = 0,1022g of La

Mass fraction of La is:

0,1022g of La÷ 0,5222g × 100 = 19,58 wt%

Grams of Ce are:

0,3417gCe(NO₃)₃×$\frac{1mol}{326,13g}$×$\frac{140,116g}{1molCe}$ = 0,1468g of Ce

Mass fraction of Ce is:

0,1468g of Ce÷ 0,5222g × 100 = 28,11 wt%

I hope it helps!