is shown e overall reaction for the synthesis of potassium iron(IlI) oxalate is

Question

is shown e overall reaction for the synthesis of potassium iron(IlI) oxalate is below 342.1088 g/mol the following (showing your calculations): lculate the number of moles in 5.00 g of Fe(NH42(SO426H2o .OV 5 9 012750 S2 mo i or.0imo! 392.14088 342.14088012750 the number of moles in 33.0 ml of 1.00 M H2C204 033 mol 100mL 33mls :33 X033 .020 he number of moles in 20.0 ml of 3% H2O2 (which is 1.00 M) l mo 1000mL5 20ml 016 he number of moles in 10.0 ml of saturated K2C2O4 (which is 2.00 M) Lmo mu 311

Explanation / Answer

As clear in the given equation the stoichiometry is:

Two moles of of ferrous ammonium sulphate reacts with three mole of oxalic acid and one mole of peroxide and three moles of potassium oxalate.

The moles of potassium oxalate = 0.01 moles ( limiting reagent)

ferrous ammonium sulphate = 0.01

oxalic acid = 0.033

peroxide = 0.02

Three moles of potasssium oxalate will give two moles of potassium iron (iii) oxalate.

So 0.01 moles of potassium oxalate will give 0.00667 moles of potassium iron (iii) oxalate

Molecular weight of potassium iron (iii) oxalate = 491.24

So amount of potassium iron (iii) oxalate formed = moles X mol wt = 0.00667 X 491.24 = 3.27 g

So the theoretical yield = 3.27g

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