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slega [8]
4 years ago
11

How many moles of sodium iodide are needed to remove 5.95×10−6 mol of o3?

Chemistry
1 answer:
igor_vitrenko [27]4 years ago
8 0
You have to find the reaction involving sodium iodide and ozone or O₃. If you search in the internet, the reaction would be

O₃<span>(g) + 2 NaI(aq) + H</span>₂O(l) → O₂<span>(g) + I</span>₂<span>(g) + 2 NaOH(aq) 
</span>
Using the stoichiometric coefficients in the reaction, the solution would be

5.95×10⁻⁶ mol O₃ * (2 mol NaI/1 mol O₃) = 1.19×10⁻⁵ mol NaI
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What is the coefficient for iodide ions (I-) when the equation is balanced?<br>​
alexgriva [62]

Answer:

When the following redox equation is balanced with smallest whole number coefficients, the coefficient for the iodide ion will be __6__.

Explanation:

From the redox equation, we can see that NO₃⁻ is reduced to NO (from oxidation state +5 to +2), whereas I⁻ is oxidized to I₂ (from oxidation state -1 to 0). The half reactions are balanced with H⁺ (acidic solution), as follows:

Reduction :        2 x (NO₃⁻(aq) + 3 e-  + 4 H⁺ → NO(g) + 2 H₂O)

Oxidation :                                   3 x (2 I⁻(aq) → I₂(s) + 2 e-)

                      ----------------------------------------------------------------------

Total equation: 6 I⁻(aq) + 2 NO₃⁻(aq)+ 8 H⁺ → 3 I₂(s) + 2 NO(g) + 4 H₂O

That is the redox equation with the smallest whole number coefficients.

Accordin to this, the coefficient for the iodide ion (I⁻) is: 6.

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