Using the balanced equation just found and the mass of fe (nh4) 2 (so4) 2·6h2o recorded at the beginning of last week's experiment determine the percent yield of kxfey (c2o4) z. nh2o.

Question

Emerson Henderson

Chemistry

20 May, 17:43

Using the balanced equation just found and the mass of fe (nh4) 2 (so4) 2·6h2o recorded at the beginning of last week's experiment determine the percent yield of kxfey (c2o4) z. nh2o.

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Rex Odom · Answer 1

The balanced equation is Fe (NH4) 2 (SO4) 2·6H2O + H2C2O4 + 2K2C2O4 + H2O2 →K4Fe1 (C2O4) 3. nH2O + (NH4) 2SO4 + H2SO4 + 8H2O

Percent Yield is actual/theoretical x 100

Actual is the 2.068g we measured after vacuum filtering in part c

Theoretical is

2.522g Fe (NH4) 2 (SO4) 2·6H2O x (1 mol Fe (NH4) 2 (SO4) 2·6H2O / 392.14g) x (1 mol K4Fe (C2O4) 3 nH2O/1mol Fe (NH4) 2 (SO4) 2·6H2O) x (346.1102 gK4Fe (C2O4) 3 nH2O / 1 mol K4Fe (C2O4) 3 nH2O) = 2.226g

Therefore percent yield is (2.068g/2.226g) x 100 = 92.9%