1. Weight of beaker (g) 2- weight of beaker and ferrous ammonium sulfate 3. Weig
ID: 553781 • Letter: 1
Question
1. Weight of beaker (g) 2- weight of beaker and ferrous ammonium sulfate 3. Weight of filter paper (g) 4· Weight of filter paper and product (g) a hexahydrate (g) Be 3-00 9 5. weight of ferrous ammonium sulfate hexahydrate (g) 6. Moles of ferrous ammonium sulfate hexahydrate (mol) 7. Predicted number of moles of x,Fe (co3Ho (mol) 8. Predicted number of grams of K3Fe (C204 )3-3H20 (g) 9. Bxperimental weight (yield) of K,Pe (Ca.3H20 (g) - 10. Percentage yield of Ka Fe (C204 )3-3Hao (%) Compute the weight of the starting material by subtracting the beaker's weight from the weight of the beaker plus ferrous ammonium sulfate hexahydrate. (Step 2 Step 1 on the Report Sheet) Report this value to two decimal places Calculate the number of moles of Fe (NH) a (SO) .6H2O present in the weight of starting material used The formula weight of the starting material Fe (NH (So 6H20 ig 392.13 g/mo Round and record your answer to 3 significant figure tific notation Determine the predicted number of moles of K,Fe (C20 3H20 (Step 7) from the number of moles of starting material used (Step 6)Round 80 Experiment #8Explanation / Answer
Preparation of K3Fe(C2O4)3.3H2O
1 mole of ferrous ammonium sulfate hexahydrate yields 1 mole of K3Fe(C2O4)3.3H2O
Calculations,
5. weight of ferrous ammonium sulfate hexahydrate = 3.0 g
6. moles of ferrous ammonium sulfate hexahydrate = 3.0 g/392.13 g/mol = 0.00765 mol
7. Predicted number of moles of K3Fe(C2O4)3.3H2O = 0.00765 mol
8. Predicted number of grams of K3Fe(C2O4)3.3H2O = 0.00765 mol x 491.243 g/mol = 3.76 g
9. Experimental number of grams of K3Fe(C2O4)3.3H2O = not given here
10. Yield of K3Fe(C2O4)3.3H2O = Experimental yield (g) x 100/3.76 g
moles = grams/molar mass
grams = moles x molar mass
%yield = (experimental yield/predicted theoretical yield) x 100
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