Question

If 0.5mL of .3% (mass) ethanol is oxidized with 10mL 0.005M K2Cr2O7 , what volume of 0.05M Fe(NH4)2(SO4)2.6H2O is needed to titrate the excess potassium dichromate?

Answer 1

CH3CH2OH + K2CrO7 = oxided ethanol + 2K+ + CrO7-2 excess

ratio between ethanol + chromate is 1:1

the reaction of excess chromate:

3 K2Cr2O7 + 10 Fe(NH4)2(SO4)2*6H2O + 14 H3SO4 = 6 CrSO4 + 5 Fe2(SO4)3 + 3 K2SO4 + 10 (NH4)2SO4 + 81 H2O

ratio is

3 mol of Cr2O7-2 per 10 mol of Fe(NH4)2(SO4)2*6H2O

then...

mass of ethanol --> 0.5 mL *0.003 g/mL = 0.0015 g of ethanol

mol of ethanol = mass/MW = 0.0015g /46 g/mol= 0.00003260 mol of ethanol = 0.00003260 mol *10^3 mmol = 0.0326 mmol of ethanol

mol of Chromate = MV = 10 mL *0.005 M= 0.05 mmol of Chromate

reaction = 1:1 ratio so

mmol difference = 0.05 - 0.0326= 0.0174 mol of dichromate left

since raito is 3:10

1 mol of chromate per 3.333 mol of Fe(NH4)2(SO4)2.6H2O

then

0.0174 *3.33 = 0.057942 mmol of Fe(NH4)2(SO4)2.6H2O is required

then

V = mmol/M =0.057942 /0.05

V = 1.15884 mL