1) An aqueous solution solution contains 1 mol of S2O3 2- ions and it reduces 4 mol of Cl2 molecules. What is the sulfur containing product of this reaction?
A] S2
B] SO3 2-
C] SO4 2-
D] S4O6 2-
2) 25.0cm3 of 0.05 mol/dm3 KClO4(aq) required 50.0cm3 of 0.20mol/dm3 TiCl3(aq) for complete reaction. Given that titanium(III) is oxidised to titanium(IV) in this reaction, which one of the following formulae correctly represents the reduction product of the ClO4 - ion?
A] Cl-
B] ClO2 -
C] ClO3 -
D] OCl -
Can explain how to solve? Thanks in advance!
Originally posted by SunnyFly:1) An aqueous solution solution contains 1 mol of S2O3 2- ions and it reduces 4 mol of Cl2 molecules. What is the sulfur containing product of this reaction?
A] S2
B] SO3 2-
C] SO4 2-
D] S4O6 2-
2) 25.0cm3 of 0.05 mol/dm3 KClO4(aq) required 50.0cm3 of 0.20mol/dm3 TiCl3(aq) for complete reaction. Given that titanium(III) is oxidised to titanium(IV) in this reaction, which one of the following formulae correctly represents the reduction product of the ClO4 - ion?
A] Cl-
B] ClO2 -
C] ClO3 -
D] OCl -
Can explain how to solve? Thanks in advance!
Q1.
4 moles of Cl2 molecules accept 8e- to be reduced to 8Cl-.
The OS of S in thiosulfate ion is +2.
Hence, each of the 2 S atoms must donate 4 e- each. Hence, the OS of S in the oxidized species must be (+2) + (+4) = +6. Hence the answer is C.
Q2.
1.25 x 10^-3 mols of perchlorate ions / chlorate(VII) ions, oxidize 0.01 mol of titanium(III) ions to titanium(IV) ions. This implies 1.25 x 10^-3 mols of perchlorate ions / chlorate(VII) ions accept 0.01 mol of e-. This implies 1 mol of perchlorate ions / chlorate(VII) ions accept 8 mol of e-. Hence the final OS of chlorine in the reduced species must be (+7) + (-8) = -1. Hence the answer is A.