Asked by Anonymous
overall balanced reaction:
14H+ + C2O7^2- + 6Fe^2+ -> 2Cr^3+ + 7H2O + 6Fe^3+
.5021g of impure sample containing potassium dichromate was analyzed by titrating a standard iron (II) sulfate solution. If 6.74 mL of 0.2312 N of iron (II) sulfate was required to reach the equivalence point, calculate the % by mass of potassium dichromate in the sample using the concept of equivalent weight. (I can't see to get the answer 15.2%)
work: .5021g/ (6.74x10^-3L)(.2312 mol/L)(6 equivalents/1L) =53.7g
I've tried dividing it by the theretical equivalent of potassium dichromate but it doesn't work.
Thanks for help!
14H+ + C2O7^2- + 6Fe^2+ -> 2Cr^3+ + 7H2O + 6Fe^3+
.5021g of impure sample containing potassium dichromate was analyzed by titrating a standard iron (II) sulfate solution. If 6.74 mL of 0.2312 N of iron (II) sulfate was required to reach the equivalence point, calculate the % by mass of potassium dichromate in the sample using the concept of equivalent weight. (I can't see to get the answer 15.2%)
work: .5021g/ (6.74x10^-3L)(.2312 mol/L)(6 equivalents/1L) =53.7g
I've tried dividing it by the theretical equivalent of potassium dichromate but it doesn't work.
Thanks for help!
Answers
Answered by
Anonymous
nevermind! I got it!
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