Question

1. 1.0678g of a standard iron(ii) ammonium sulfate, (NH4)2Fe(SO4)2. 6H202 requires 35.50mL of permanganate solution for titration. calculate the molarity of the permanganate solution.

2. if 2.0500g of iron unknown required 30.00mL of the above permanganate solution. Calculate the % Fe in the sample.

Answer 1

1) reaction si

5 Fe²⁺ + MnO₄^- + 8 H⁺...............> 5 Fe³⁺ + Mn²⁺ + 4 H₂O .

5 mole Fe2+ react with 1 mole permanganate ion.

1.0678 g of a standard iron(ii) ammonium sulfate, (NH4)2Fe(SO4)2. 6H20 = 1.0678 g / 392.13 g/mole = 2.723 * 10^-3 mole.

thus

mole of permanganate = 2.723 * 10^-3 mole / 5 = 5.446 * 10^-4 mole.

molarity of the permanganate solution = 5.446 * 10^-4 mole / 0.03550 L = 0.01534 M (answer)

2) 30.00 ml of 0.01534 M permanganate = 0.030 L * 0.01534 mole / L = 4.6024 * 10^-4 mole.

mole of Fe = 5 * 4.6024 * 10^-4 mole = 2.30 * 10^-3 mole

mass of Fe in sample = 2.30 * 10^-3 mole * 55.85 g / mole = 0.1285 g

% of Fe = 0.1285 * 100 / 2.0500 = 6.27 % (answer)