Question

A reaction of 7.20 mL of Iron(III) chloride solution (0.40 g/mL) and 8.95 g of K₂C₂O₄.H₂O (Potassium oxalate monohydrate) produced 6.22 g of K₃Fe(C₂O₄)₃.3H₂O [Potassium trioxalatoferrate(III) trihydrate]

A) Calculate number moles of Iron(III) chloride?

B)Calculate number moles K₂C₂O₄.H₂O (Potassium oxalate monohydrate).

C)What is limiting reagent of the reaction (Consider stochiometry)?

D)Molar mass of Potassium trioxalatoferrate(III) trihydrate

E)Calculate the percent yield of the synthesis?

Answer 1

Balanced reaction given below
FeCl3 + 3K2C2O4·H2O ---> K3(Fe(C2O4)3)·3H2O + 3KCl

given

volume of FeCl3 = 7.2 ml
density = 0.4 g/ml
SO mass of FeCL3 present = density *volume= 7.2 ml*0.4 g/ml
= 2.88 g

molar mass of FeCL3 =1*55.84 g/mol + 3*35.45 g/mol= 162.2 g/mol
Number of moles present = mass/molar mass= 2.88g/162.2 g/mol =0.0177 mols

B)
given mass of K2C2O4.H2O = 8.95 g

molar mass of K2C2O4.H2O = [2*39.10 + 2*12.01+5*16 + 2*1.01} g/mol
= 184.23 g/mol

Number of mols of K2C2O4.H2Opresent = 8.95 g/184.23 g/mol =0.04858 mols

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C) 1 mole FeCl3 need 3 mol K2C2O4 H2O
So 0.0177 molFeCl3 need 3* 0.0177 mol K2C2O4 H2O = 0.0531
But moles ofK2C2O4 H2O =0.04858 mols
which is fewer .Hence K2C2O4 H2O act as limiting reactant.The reaction will stop once
all K2C2O4 H2O is consumed
limiting reactant =K2C2O4 H2O

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D)
molar mass of K3(Fe(C2O4)3)·3H2O
=[3*39.10 +1*55.84+6*12.01+15*16+6*1.01 = 491.26 g/mol

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limiting reactant =K2C2O4 H2O so it limits the reaction. It decide
amount of product formed
we have mole ratio K2C2O4 H2O :K3Fe(C2O4)3.3H2O = 3:1

SO moles of K3Fe(C2O4)3.3H2O produced = 1/3*0.04858 mols = 0.01619 mols
mass of K3Fe(C2O4)3.3H2O produced = number of moles of K3Fe(C2O4)3.3H2O *molar mass of
K3Fe(C2O4)3.3H2O = 0.01619 mols*491.26 g/mol =7.95 g

This is the theoretical yield ofK3Fe(C2O4)3.3H2O
actual yield is 6.22 g

Percent yield = actual yield/theoretical yield *100 = 6.22/7.95 *100 = 78.24%
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hope it is helpful