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

1 answer:

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(g) + 2 NaI(aq) + H₂O(l) → O₂(g) + I₂(g) + 2 NaOH(aq)

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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Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that c

malfutka [58]

Answer:

$[F^-]_{max}=4x10{-3}\frac{molF^-}{L}$

Explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

$Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)$

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

$[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\$

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