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I am very lost on how to combine these two equations to get a single equation.?
What is the ratio of moles of IO3– to moles of S2O32– for the titration you will be performing in this experiment? [HINT: Rewrite equations 4 & 5 from the lab directions to give one overall reaction with the I3– canceling out completely (since it is being reacted away).]
1 Answer
- Roger the MoleLv 75 days ago
Add up everything on the left sides of both arrows in the given equations and place the sum on the left side of a new arrow:
IO3{-} + 8 I{-} + 6 H{+} + I3{-} + 2 S2O3{2-} →
Now add up everything on the right sides of both arrows in the given equations and place the sum on the right side of the new arrow:
IO3{-} + 8 I{-} + 6 H{+} + I3{-} + 2 S2O3{2-} → 3 I3{-} + 3 H2O(ℓ) + 3 I{-} + S4O6{2-}
Cancel everything that appears on both sides of the new arrow. (In this case only I{-} and I3{-} appear on both sides of the equation, so subtract the lesser amount of I{-} (3 mol) and I3{-} (1 mol) from each side of the equation.) That leaves:
IO3{-} + 5 I{-} + 6 H{+} + 2 S2O3{2-} → 2 I3{-} + 3 H2O(ℓ) + S4O6{2-}
From this last equation we see that the molar ratio of IO3{-} to S2O3{2-} is 1 : 2 .
