[DUDE778]

The concentration of standardized EDTA is determined by titration of 25.0 mL of 0.100 M CaCl2 solution by 50.0 mL EDTA.

Aluminum is determined by titration of Al3+ with the standard EDTA.

Al3+ + H2Y2− ® AlY− + 2H+

A 1.00-g sample is titrated by 34.0 mL EDTA.
Calculate the percent Al2O3 in the sample.

I just want to make sure i am right..
0.100 M CaCl2 x 0.025 L = 0.0025 CaCl2 x 1 mol EDTA / 1 mol CaCl2 = 0.0025 mol EDTA
0.0025 mol EDTA / 0.05 L = 0.05 M EDTA
0.05 M EDTA x 0.034 L = 0.0017 mol EDTA x 1 mol AlY- / 1 mol EDTA = 0.0017 mol AlY-
0.0017 mol AlY- x 101.961 g Al2O3 / 1 mol AlY- = 0.1733 g Al2O3
0.1733 g / 1.00 g x100% = 17.3 %

Is this write    I think i might have got lost when i had 0.0017 mol AlY- and tried to get to Al2O3

[JGK]

You neet to balance this equation properly:

Al³⁺ + H2Y2− ® AlY− + 2H⁺

     

the charges on each side dont match!

[AWK]

Al³⁺ + H₂Y²⁻ ® AlY⁻ + 2H⁺

Equation is corect (with some printing errors)