the mass of H2C2O4 acid in a solid sample

Please see the attached file for detailed solution.

neutralization reaction
oxalic acid H2C2O4 is a diprotic acid. calculate the mass of acid in a solid given that a 0.7984 g sample of that solid required 45.00 mL of 0.300 M NaOH for neutralization.

1 mol of H2C2O4 can provide 2 moles of H+ in the neutralization since it is a diprotic acid.

For NaOH, 1 mole of NaOH only provide 1 mol of OH-.

H+ + OH-  H2O
In the above reaction, 1 mole of H+ reacts with 1mol of OH .
Therefore, in the neutralization, 1 mol of H2C2O4 will react with 2 mol of NaOH.
That is, the moles of NaOH is twice that of H2C2O4 in the reaction.

The moles of NaOH used is
(45.00 mL) * (0.300 M) = (0.0450 L) * (0.300 M) = 0.0135 mol.

Then it requires (0.0135 mol) / 2 = 0.00675 mol of H2C2O4 in neutralization.

The molar mass of H2C2O4 is 2 *1 + 2* 12 + 4* 16 = 90 g/mol
So the mass of 0.00675 mol of H2C2O4 is
(90 g/mol) * (0.00675 mol) = 0.6075 grams.

There are 0.6075 g of oxalic acid in the solid.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!