Calculate M for trial 1 and 2 show calculation Weight of Fas: Trial 1-1.00g Trial 2-...

Calculate M for trial 1 and 2 show calculation

Weight of Fas: Trial 1-1.00g Trial 2- 1.00 g

Final reading of buret: 18.00 mL 18.00mL

Initial reading of buret: 50.00 mL 50.00mL

V to equivalence pt.: 26.00mL 26.00mL

Molarity of KMnO4: ? ?

Solutions

Expert Solution

stoichiometric equation

2 KMnO4(aq) + 10(NH4)2Fe(SO4)2(aq) + 8H2SO4(aq) -------> 2MnSO4(aq) + 5Fe2(SO4)3(aq) + 10(NH4)2SO4(aq) + K2SO4(aq) + 8H2O

according stoichiometry 2 mole of KMnO4 react with 5 mole of FAS ( the ratio of KMnO4/FAS is 1:5)

Now,

Trial 1

Mazs of FAS = 1.00g

Molar mass of FAS = 392.13g/mol

No of mole of FAS = 1g/392.13g = 0.00255

So , No of mole of KMnO4 reacted = 0.00255/5=0.00051

Volume of FAS consumed = 26ml

Molarity of FAS = (0.00051/26ml)×1000ml = 0.0196M

Trial 2

Volume of FAS consumed

Molarity of FAS = 0.0196M

Note: As per Initial and final burette reading , volume of KMnO4 consumed is 32ml , if it is true Molarity of KMnO4 is 0.0159M for both trials

queen_honey_blossom answered 1 year ago

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