This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:
Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰
Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻
Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.
In such a way, the correct choice is C.
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Answer:
9.6 moles O2
Explanation:
I'll assume it is 345 grams, not gratis, of water. Hydrogen's molar mass is 1.01, not 101.
The molar mass of water is 18.0 grams/mole.
Therefore: (345g)/(18.0 g/mole) = 19.17 or 19.2 moles water (3 sig figs).
The balanced equation states that: 2H20 ⇒ 2H2 +02
It promises that we'll get 1 mole of oxygen for every 2 moles of H2O, a molar ratio of 1/2.
get (1 mole O2/2 moles H2O)*(19.2 moles H2O) or 9.6 moles O2
