Using the correct formula for the green crystals write a balanced chemical equation for their synthesis....
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Using the correct formula for the green crystals write a balanced chemical equation for their synthesis. The reactants were iron (III) chloride, potassium oxalate and water. A second product of the synthesis is potassium chloride. Note: the waters of hydration in the iron(III) chloride, and potassium oxalate should not be shown. Include phase labels.
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Using the masses of iron (III) chloride hexahydrate and potassium oxalate monohydrate from Data Table A, calculate the theoretical yield of green crystals. This is a limiting reactant problem. Show all your work here.(My data: FeCl3*6H2O is 3.662g; K2C2O4*H2O is 8.226g.)
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Using your experimental mass of crystals and your answer to the previous question, calculate the percent yield for your synthesis. Show your work here. (My experimental mass is 2.888g)
4. Write the net-ionic chemical equation for the reaction that produced the green crystals. Hint: Work backwards from the product. Include phase labels.
Homework Answers
FeCl3(aq) + 3K2C2O4(aq) + 3H2O(l) ---> K3 [Fe(C2O4)3 . 3H2O (s) + 3KCl (aq)
Green crystals
Total ionic equation :
Fe+3 (Aq) + 3Cl-(aq) + 6K+ (aq) + 3C2O4-2 (aq) + 3H+(aq) + 3OH-(aq) ---> K3 [Fe(C2O4)3 . 3H2O (s) + 3K+ (aq)+3Cl-(aq)
Net ionic equation :
Fe+3 (Aq) + 3K+ (aq) + 3C2O4-2 (aq) + 3H+(aq) + 3OH-(aq) ---> K3 [Fe(C2O4)3 . 3H2O (s)
FeCl3.6H2O (aq) + 3K2C2O4 H2O(aq) ---> K3 [Fe(C2O4)3 . 3H2O (s) + 3KCl (aq) + 3H2O(l)
Mass of FeCl3.6H2O = 3.662 g
Molar mass of FeCl3.6H2O = 270.3 g/mol
No. of moles of FeCl3.6H2O = Mass/Molar mass = 3.662 g/270.3 g/mol = 0.0135 moles
Molar ratio of FeCl3.6H2O and K3 [Fe(C2O4)3 . 3H2O is 1:1
So, moles of K3 [Fe(C2O4)3 . 3H2O is 0.0135 moles -------- 1
Mass of K2C2O4 H2O = 8.226 g
Molar mass of K2C2O4 H2O = 184.2 g/mol
No. of moles of K2C2O4 H2O = Mass/molar mass = 8.226 g/184.2 g/mol = 0.0447 moles
Molar ratio of K2C2O4 H2O and K3 [Fe(C2O4)3 . 3H2O is 3:1
So, moles of K3 [Fe(C2O4)3 . 3H2O is 0.0447 moles/3 = 0.0149 moles -------- 2
From 1 and 2 we can observe that FeCl3.6H2O is limiting reactant
Molar mass of K3 [Fe(C2O4)3 . 3H2O = 491.2 g/mol
So, theoretical mass of K3 [Fe(C2O4)3 . 3H2O = theoretical moles x molar mass = 0.0135 moles x 491.2 g/mol = 6.631 g
Experimental yield = 2.888 g
Percent yield = (Experimental yield/theoretical yield) x 100 = (2.888 g/6.631 g) x 100 = 43.55 %
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