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 .

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%

13) a 14) a15) b16) i don't know sorry17) a (maybe)