Asked by Tichele
36.85 mL of a KMnO4 solution when acidified reacted with 1.076g of Mohr's salt, Fe(NH4)2(SO4)2 x 6H2O What is the normality of the KMnO4 solution?
How do I set this up? I can't figure it out at all...
How do I set this up? I can't figure it out at all...
Answers
Answered by
DrBob222
MnO4- + 5Fe2+ ==> Fe3+ + Mn2+
N = #milliequivalents/mL = #me/mL.
#me = grams/mew (where mew = milliequivalent weight)
#me = g/mew = 1.076/0.39214 = 2.7439
N = me/mL = 2.7439/36.85 = ??
Note: mew of Mohr's salt = molar mass/delta e. 5 moles Fe undergo 5e change which is 1e/1 mol so molar mass = equivalent mass (equivalent weight).
N = #milliequivalents/mL = #me/mL.
#me = grams/mew (where mew = milliequivalent weight)
#me = g/mew = 1.076/0.39214 = 2.7439
N = me/mL = 2.7439/36.85 = ??
Note: mew of Mohr's salt = molar mass/delta e. 5 moles Fe undergo 5e change which is 1e/1 mol so molar mass = equivalent mass (equivalent weight).
Answered by
jarod
no solution
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