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0.0032 74001L 12 184.32 =0.065sin actual: yiud Two students did this experiment using 8 mL 12.0 g of K2C2O4·H2O. They recovered 7.126 g of 0.40 (g)/(mL) FeCl solution and a. What is their actual yield? theoretical = 1:3 molar yield 0.40:1.2, 0.05 * (491.2497 g) = 24.562485 c. What is their percent yield? % yield = ((actual)/(theoretical)) * 100 = ((7.126)/(24.562485 g)) * 100 = 29% 29090 11600 12 184.32 3. Two students did this experiment using 8 mL of 0.40 g/mL FeCl solution and 12.0 g of K2C2O4·H2O. They recovered 7.126 g of K3[Fe(C2O4)3]·3H2O. o l Imoi 91.24979 mol 7.12 g 0.40 mg 0.00% b. What is their theoretical yield? ratio yield 0.40:1.2 0.05 * 491.2497 g = 24.607485 c. What is their percent yield? 20.9 001.7 - 1269 10012910 theoretical

          0.0032
74001L
12
184.32
=0.065sin
actual:
yiud
Two students did this experiment using 8 mL
12.0 g of K2C2O4·H2O. They recovered 7.126 g of 0.40 (g)/(mL) FeCl solution and
a. What is their actual yield?
theoretical = 1:3 molar
yield 0.40:1.2, 0.05 * (491.2497 g) = 24.562485
c. What is their percent yield?
% yield = ((actual)/(theoretical)) * 100 = ((7.126)/(24.562485 g)) * 100 = 29%
29090
11600
12 184.32
3. Two students did this experiment using 8 mL of 0.40 g/mL FeCl solution and 12.0 g of K2C2O4·H2O. They recovered 7.126 g of K3[Fe(C2O4)3]·3H2O.
o l Imoi 91.24979 mol 7.12 g 0.40 mg 0.00%
b. What is their theoretical yield?
ratio yield 0.40:1.2 0.05 * 491.2497 g = 24.607485
c. What is their percent yield? 20.9 001.7 - 1269 10012910 theoretical

00032 74001l 12 18432 0065sin actual yiud two students did this experiment using 8 ml 120 g of k2c2o4h2o they recovered 7126 g of 040 gml fecl solution and a what is their actual yield theor 48888

Added by Andrea C.

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