a) Find the empirical formula for KxFey(C2O4)z·nH2O given % percents:
potassium = 34.5%
iron = 11.4%
water = 11.0%
oxalate = 43.1 %
Take a 100 g sample. That will give you
34.5 g K
11.4 g Fe
11.0 g H2O
43.1 g oxalate (C2O4)
Now divide mols = g/molar mass.
From mols of each, determine the ratio of one to the other. The easy way to do this is to divide the smallest # mols by itself. That will make that element/polyatomic ion 1.000. Then, to keep things equal, divide the other mols by that same number, then round to the enarest whole number (but I caution not to round numbers from 0.4 to 0.6 up or down). That will give you the formula. Do that part, then we can talk about the remainder. You may be able to solve the rest of the problem after getting the formula.
a) Find the empirical formula for KxFey(C2O4)z·nH2O given % percents:
potassium = 34.5%
iron = 11.4%
water = 11.0%
oxalate = 43.1 %
b) Insert values for x, y, z into KxFey(C2O4)z·nH2O then give correct coefficients to balance the equation:
_Fe(NH4)2(SO4)2*6H20 + __H2C2O4 + __K2C2O4 + H2O2 --> __K3Fe(C2O4)3*3H20 + __(NH4)2SO4 + __H2SO4 + H20
I have this part, I know that it's
2 Fe(NH4)2(SO4)2*6H20 + 3 H2C2O4 + 3 K2C2O4 + 1 H2O2 --> 2 K3Fe(C2O4)3*3H20 + 2 (NH4)2SO4 + 2 H2SO4 + 2 H20
c) Using the balanced equation and mass of Fe(NH4)2(SO4)2*6H20 (2.501g) determine the percent yield of KxFey(C2O4)z·nH2O.
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