Question

Hello! I was wondering if anyone can check my work on my prelab calculation as I feel iffy about it. the question asks "calculate the mass of H2C2O4•2H2O you will require to react with about 16 ml of your potassium permangate solution". So I used the moles of our solution which was 0.02 as a conversion factor to get to the grams of H2C2O4, is this correct

or did I do something wrong?

1122 4 + 2 KMnO4 8 CO₂ + ko + MnO₂ + 4H2O with mass of H₂COy neaquired to react → 0.02 KMnO 4 m "calwiche mass of H₂ 2 about 16 mi of permanganele solution mm = 126.08g MLL KMnOy - KM nuy morning mot ou 9 H₂C2O4 H₂ C₂04 16.m. Of Helly I kenoy 0.02 monoy 4 mol H₂604 126.08g. KMnoy x -0.08009 103m. zmol Imol Kindy Vithet * 1 sig fig H₂ 204 UUUU = 0.08g of fizlou 111TTI

Answer 1

Calculate the mass of H₂C₂O₄•2H₂O will be required to react with about 16 ml of your potassium permangate solution".

KMnO₄ solution = 0.02 Mole / L  

As per Reaction shown by you :

2 KMnO₄ + 4 H₂C₂O₄ $\rightarrow$ Mn₂O₃ +8 CO₂ + 4 H₂O + K₂O

2 moles of KMnO₄ reacts with 4 moles of oxalic acid.

(you check reaction provided by you, as per redox reaction between these two : 2 moles of KMnO₄ reacts with 5 moles of oxalic acid. )

Calculations :

Moles of KMnO₄ involved = (16 ml * 0.02 mol /L ) / (1000 ml /L ) = 3.2*10^-4 moles

Since, as Reaction shown by you 2 moles of KMnO₄ reacts with 4 moles of oxalic acid,

we will need = 2*3.2*10^-4 moles = 6.4*10^-4 moles of H₂C₂O₄•2H₂O.

Molar mass of H₂C₂O₄•2H₂O = 126.08 g / mol

So, mass of H₂C₂O₄•2H₂O required = 126 g / mol * 6.4*10^-4 mole = 0.08069 g

(You are correct , if you are taking correct reaction into consideration )