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 = + <em>(1)</em>
As total mass of the solid sample are 0,5222g:
0,5222g = X + Y <em>(2)</em>
Replacing (2) in (1):
Y = 0,3417g
Thus:
X = 0,1805g
Grams of La are:
0,1805gLaCl₃×× = 0,1022g of La
Mass fraction of La is:
0,1022g of La÷ 0,5222g × 100 = <em>19,58 wt%</em>
Grams of Ce are:
0,3417gCe(NO₃)₃×× = 0,1468g of Ce
Mass fraction of Ce is:
0,1468g of Ce÷ 0,5222g × 100 = <em>28,11 wt%</em>
I hope it helps!