Titration Question help!

H2C2O4‧H2O + 2NaOH ---> Na2C2O4 + 3H2O --- ( 1 )
NaOH + HCl ---> NaCl + H2O --- ( 2 )
No. of moles of oxalic acid in 250ml solution:
1.620 / ( 1.0 x 2 + 12.0 x 2 + 16.0 x 4 + 1.0 x 2 + 16.0 )
= 0.015mol
No. of moles of oxalic acid in ( 1 ): 0.015 x 25 / 250 = 1.5 x 10-3 mol
No. of moles of NaOH in ( 1 ): 1.5 x 10-3 x 2 = 3 x 10-3 mol
Molarity of NaOH: 3 x 10-3 / ( 22.73 / 1000 ) = 0.132M
No. of moles of NaOH in ( 2 ): 0.132 x 12.65 / 1000 = 1.67 x 10-3 mol
Molarity of HCl in ( 2 ): 1.67 x 10-3 / ( 10.00 / 1000 ) = 0.167M

2009-02-22 19:19:15 補充：
Correction: No. of moles of oxalic acid:
1.62 / ( 2 + 12 x 2 + 16 x 4 + 2 x 18 ) = 0.0129
No. of moles of oxalic acid in 25ml: 0.0129 x 25 / 250 = 1.29 x 10^-3
No. of moles of NaOH: 1.29 x 10^-3 x 2 = 2.57 x 10^-3

2009-02-22 19:19:27 補充：
Molarity of NaOH: 2.57 x 10^-3 / ( 22.73 / 1000 ) = 0.113M
No. of moles NaOH in 2: 0.113 x 12.65 / 1000 = 1.43 x 10^-3
Molarity of HCl: 1.43 x 10^-3 / ( 10 / 1000 ) = 0.143M

2009-02-22 19:21:25 補充：
Thx for reminding. I didn't realize it while I was doing the calculation.
But jackiii_iii please kindly check your Q before you post it=]

The formula of oxalic acid dihydrate should be H2C2O4•2H2O instead of H2C2O4•H2O. Due to this mistake in the question, Gabriella Montez got a small mistake in calculating the molar mass of the acid.