The students used 90.0 g/mol as the molar mass of oxalic acid. Use this value to determine how the students calculated the number of moles of oxalic acid present that were completely neutralized by the base.

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Explanation:

Oxalic acid is a diprotic acid and the acid is used with or react with sodium Hydroxide, NaOH in order to determine the molar mass of unknown diprotic acid. Oxalic Acid reacts with sodium as it is given in the balanced chemical reaction below:

(COOH)₂ + 2NaOH > Na₂C₂O₄ + 2H₂O.

The amount of the oxalic needed is known, say x gram and the molar mass is known. Thus, the number of moles of the oxalic acid= mass/molar mass = x gram/ 90.0 g/mol.

So we say that (COOH)₂ is a primary standard acid which is been used to standardize the base that is Sodium hydroxide.

Then, if we have an unknown diprotic acid, say H₂A, we can react it with NaOH to get the molar mass of the unknown.

H₂A + 2NaOH > Na₂A + 2H₂O.

Therefore, the number of moles of the acid = [concentration of NaOH × volume of NaOH] × [ 1 mole of the acid/ 2 mole of NaOH].

if you mean hydrogen and oxygen then water is the product

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