How do you balance $H_2C_2O_4 + NaOH -> Na_2C_2O_4 + H_2O$ ?

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Answer 1 · 2016-01-15T18:01:12

Oxalic acid + sodium hydroxide $rarr$ sodium oxalate + water

Oxalic acid, $HO(O=)C-C(=O)OH$ , can react with 2 equiv base in that there are 2 carboxylic acid functions:

$HO(O=)C-C(=O)OH(aq) + 2NaOH(aq) rarr {C(=O)O^-}_2Na_2^+ + 2H_2O(l)$

Is this balanced? For every reactant particle is there a corresponding product particle?

How do you balance H_2C_2O_4 + NaOH -> Na_2C_2O_4 + H_2OH_2C_2O_4 + NaOH -> Na_2C_2O_4 + H_2O?