Consider the following net ionic equation. Which species is oxidized? MnO_4^- +
ID: 1010568 • Letter: C
Question
Consider the following net ionic equation. Which species is oxidized? MnO_4^- + SO_3^2- rightarrow Mn^2+ + SO_4^2- (acidic solution) MnO_4^- Mn^2+ SO_3^2- SO_4^2- no species is oxidized Balance the following net ionic equation. What is the sum of the coefficients? H_2CO + AlH_4^- rightarrow CH_3OH + Al^3+ (in basic solution) 26 16 22 18 13 What volume of 0.230 M potassium hydroxide solution, KOH, would completely neutralize 27.4 mL of 0.205 M sulfurous acid solution, H_2SO_3? 61.5 mL 12.2 mL 24.4 mL 48.8 mLExplanation / Answer
1a) in MnO4- oxidation state of Mn = +7
in Mn+2 oxidation state of Mn = +2
means Mn changes from +7 to +2
decreasing oxidation number so MnO4- is reduced
in SO3-2 S = +4 oxidation number
answer = option C = SO3-2
in SO4-2 S = +6 oxidation number
so increasing oxidation state
SO3-2 is oxidized
1 b ) 4 H2CO + AlH4- + 4 H2O --------> 4CH3OH + Al+3 + 4OH-
4+1+4+4+1+4 = 18
answer = option d) 18
1c) H2SO3 + 2KOH --------------> K2SO3 + 2H2O
moles of KOH = 0.205 x 0.0274 = 0.005617
according to balanced reaction
2 moles KOH reacts with 1 mole of H2SO3
0.005617 mole KOH require 0.005617 / 2 =0.0028085 moles H2SO4
moles = Molarity x volume in liters
0.0028085 = 0.230 x V
V = 0.0122 L = 12.2 mL
answer = option b = 12.2 mL