Question

3. A fertiliser contains ammonium iron (II) sulphate, FeSO4.(NH4)2SO4.6H2O as a source of iron. A 6.50 g sample of this fertiliser is made up to 250 mL with dilute sulphuric acid. 25 mL of this solution is reacted with 23.5 mL of 0.01 mol L-1 potassium dichromate (K2Cr2O7) solution as shown by the balanced equation below.

Cr2O7²- + 6Fe²+ +14H+ ---> 2Cr³+ + 6Fe³+ + 7H2O

(i) Calculate the moles of Cr2O72+ ions that has reacted. 

(ii) Determine the moles of Fe2+ ions that has reacted

(iii) Calculate the mass of Fe in the 25 mL aliquot (M(Fe) = 56 g mol-1))

(iv) Determine the total mass of Fe present in the 6.50 g sample of fertiliser. 

(v) Calculate the percentage of Fe in the fertiliser. 

Answer 1

Question:

3. A fertiliser contains ammonium iron (II) sulphate, FeSO4.(NH4)2SO4.6H2O as a source of iron. A 6.50 g sample of this fertiliser is made up to 250 mL with dilute sulphuric acid. 25 mL of this solution is reacted with 23.5 mL of 0.01 mol L-1 potassium dichromate (K2Cr2O7) solution as shown by the balanced equation below.

Cr2O7²- + 6Fe²+ +14H+ ---> 2Cr³+ + 6Fe³+ + 7H2O

(i) Calculate the moles of Cr2O72+ ions that has reacted. 

(ii) Determine the moles of Fe2+ ions that has reacted

(iii) Calculate the mass of Fe in the 25 mL aliquot (M(Fe) = 56 g mol-1))

(iv) Determine the total mass of Fe present in the 6.50 g sample of fertiliser. 

(v) Calculate the percentage of Fe in the fertiliser. 

Answer:

(i) The number of the moles of Cr₂O₇^2- ions can be calculated as the product of the concentration and the volume of the solution:

$n = cV = 0.01·0.0235 = 0.000235$ mol.

(ii) According to the reaction equation, 6 moles of iron ions react with 1 mole of dichromate:

$n(Fe^{2+}) = 6n(Cr_2O_7^{2-}) = 6·0.000235 = 0.00141$ mol.

(iii) The mass of Fe is the product of its number of the moles and its molar mass:

$m(Fe) = n(Fe)·M(Fe) = 0.00141·56 = 0.079$ g.

(iv) The total mass of Fe is the mass in the aliquot multiplied by a ratio of the total volume of the solution to the volume of the aliquot:

$m(Fe)_{tot} = m(Fe)_{aliquot}·\frac{V_{total}}{V_{aliquot}} = 0.079·\frac{250}{25} = 0.79$ g.

(v) The percentage of Fe in the fertiliser is:

$w(Fe) = \frac{m(Fe)}{m(sample)} ·100\% = \frac{0.79}{6.50} ·100\% = 1.2\%$ .