Question

When a 0.80 g sample of unknown Fe2+ salt is titrated with 0.0285 M K2Cr2O7. it is found that 59.2 mL of this dichromate solution are required to reach the equivalence endpoint b. calculate the percent iron in the unknown using the balanced equation

Answer 1

Answer: The percent of iron (II) ions in the sample is 71.25 %

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of dichromate solution = 0.0285 M

Volume of solution = 59.2 mL = 0.0592 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.0285M=\frac{\text{Moles of }K_2Cr_2O_7}{0.0592L}\\\\\text{Moles of }K_2Cr_2O_7=(0.0285mol/L\times 0.0592L)=0.0017mol$

The chemical equation for the reaction of iron (II) ions with potassium dichromate follows:

$Cr_2O_7^{2-}+6Fe^{2+}+14H^+\rightarrow 2Cr^{3+ }+6Fe^{3+}+7H_2O$

By Stoichiometry of the reaction:

1 mole of dichromate solution reacts with 6 moles of iron (II) ions

So, 0.0017 moles of dichromte solution will react with = $\frac{6}{1}\times 0.0017=0.0102mol$ of iron (II) ions

• To calculate the mass of iron (II) ions for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of iron = 55.85 g/mol

Moles of iron = 0.0102 moles

Putting values in above equation, we get:

$0.0102mol=\frac{\text{Mass of iron (II) ions}}{55.85g/mol}\\\\\text{Mass of iron (II) ions}=(0.0102mol\times 55.85g/mol)=0.570g$

• To calculate the mass percentage of iron (II) ions in sample, we use the equation:

$\text{Mass percent of iron (II) ions}=\frac{\text{Mass of iron (II) ions}}{\text{Mass of sample}}\times 100$

Mass of sample = 0.80 g

Mass of iron (II) ions = 0.570 g

Putting values in above equation, we get:

$\text{Mass percent of iron (II) ions}=\frac{0.570g}{0.80g}\times 100=71.25\%$

Hence, the percent of iron (II) ions in the sample is 71.25 %