Question

2. A mineral specimen is analyzed to determine the amount of cassiterite (SnO2) it contains. A 0.5316 g sample of the mineral is prepared by dissolution of the solid into an acid solution. The Sn4+ cations in the solution are reduced to Sn2+. The solution is titrated with 0.01804 M K2Cr2O7 requiring 22.96 mL to reach the end point. The net ionic equation for the titration is given below.

3 Sn2+ (aq) + Cr2O72– (aq) + 14 H+ (aq) → 3 Sn4+ (aq) + 2 Cr3+ (aq) + 7 H2O (l)

a) Identify the oxidizing agent and the reducing agent in the titration reaction. (3 pts)

b) Determine the %SnO2 in the mineral sample. (3 pts)

Answer 1

+2 +6 +4 +3

3 Sn2+ (aq) + Cr2O72– (aq) + 14 H+ (aq) → 3 Sn4+ (aq) + 2 Cr3+ (aq) + 7 H2O (l)

Here Sn2+ decreases the oxidation state of Cr in Cr2O7 2- from +6 to +3 in Cr3+ so Sn2+ is the reducing agent.

Also  Cr2O72– increases the oxidation state of Sn in Sn2+ from +2 to +4 in Sn4+ so  Cr2O72– is the oxidizing agent.

The number if moles of K2Cr2O7 , n = Molarity x volume in L

= 0.01804 M x 22.96 mL x 10^-3 L/mL

= 4.14x10^-4 mol

From the balanced equation ,

1 mole of K2Cr2O7 reacts with 3 moles of Sn2+

4.14x10^-4 mol of K2Cr2O7 reacts with 3*4.14x10^-4 = 1.24x10^-3 moles of Sn2+

Mass of SnO2 , m = number of moles x molar mass

= 1.24x10^-3 mol x 150.7 g/mol

= 0.187 g

% SnO2 = ( mass of SnO2 / mass of sample ) *100

= ( 0.187 g / 0.5316g) *100

= 35.2%