Question

H2C2O4 + 2NaOH -> Na2C2O4 + 2H2O

if n(H₂C₂O₄*2H₂O) = 6.44*10^-6, what is the amount of NaOH in moles?

Answer 1

H₂C₂O₄ + 2NaOH $\rightarrow$  Na₂C₂O₄ + 2H₂O

According to the stoichiometry of the reaction, the number of the moles of the oxalic acid relates to the number of the moles of sodium hydroxide as:

$n(H_2C_2O_4) = \frac{n(NaOH)}{2}$ .

The number of the moles of H₂C₂O₄ equals the number of the moles of H₂C₂O₄*2H₂O, as there is one molecule of oxalic acid per hydrate unit.

Therefore, if $n(H_2C_2O_4*2H_2O) = 6.44·10^{-6}$ mol, the amount of NaOH in moles is $n(NaOH) = 2·n(H_2C_2O_4*2H_2O) = 1.29·10^{-5}$ mol.