A sample of a tin ore weighing 0.2757 g was dissolved in an acid solution and al
ID: 790569 • Letter: A
Question
A sample of a tin ore weighing 0.2757 g was dissolved in an acid solution and all the tin in the sample was changed to tin(II). In a titration, 8.42 mL of 0.0407 M KMnO4 solution was required to oxidize the tin(II) to tin(IV).
(a) What is the balanced net ionic equation for the reaction in the titration?
(b) How many grams of tin were in the sample?
(c) What was the percentage by mass of tin in the sample?
(d) If the tin in the sample had been present in the compound SnO2, what would have been the percentage by mass of SnO2 in the sample?
I would greatly appreciate a step by step explanation of this problem, please.
Explanation / Answer
(a) Oxidation: Sn2+(aq) => Sn4+(aq) + 2 e-
Reduction: MnO4-(aq) + 8 H+(aq) + 5 e- => Mn2+(aq) + 4 H2O(l)
Add 5 x Oxidation + 2 x Reduction and cancel common terms to get net ionic equation:
2 MnO4-(aq) + 5 Sn2+(aq) + 16 H+(aq) => 2 Mn2+(aq) + 5 Sn4+(aq) + 8 H2O(l)
(b) Moles of MnO4- = volume x concentration of KMnO4
= 8.42/1000 x 0.0407 = 0.000342694 mol
Moles of Sn = 5/2 x moles of MnO4-
= 5/2 x 0.000342694 = 0.000856735 mol
Mass of Sn = moles x molar mass of Sn
= 0.000856735 x 118.71
= 0.1017 g (or approximately 0.102 g)
(c) %Mass of Sn = mass of Sn/mass of sample x 100%
= 0.1017/0.2757 x 100%
= 36.89% (or approximately 36.9%)
(d) Moles of SnO2 = moles of Sn = 0.000856735 mol
Mass of SnO2 = moles x molar mass of SnO2
= 0.000856735 x 150.71 = 0.1291 g
%Mass of SnO2 = mass of SnO2/mass of sample x 100%
= 0.1291/0.2757 x 100%
= 46.83% (or approximately 46.8%)