If 0.9829 g of oxalic acid dihydrate are to be used in the reaction and the conc

Question

If 0.9829 g of oxalic acid dihydrate are to be used in the reaction and the concentration of the stock NaOH solution is 1.0017 mol/kg, what mass of the NaOH solution should be used to have 5% excess NaOH? The formula weight of oxalic acid dihydrate is 126.04 g/mol, and the formula weight of NaOH is 40.00 g/mol.

Explanation / Answer

1mole oxalic acid + 2 moles NaOH ----> water + disodium oxalate, 0.9829 gm oxalic acid = 0.9829/126.04 = 0.0078 moles , moles of NaOH required = 2 x0.0078 = 0.0156 moles , for 5% extra we require = (0.0156 + (5 x0.0156/100)) = 0.01638 moles, M= 1.0017 moles /kg , so mass of NaOH required = (0.01638 x 1/1.0017) = 0.01635 kg = 16.35 gm NaOH required.

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