Question

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the sample through a "bubbler" containing sodium iodide, which removes the ozone according to the chemical equation: O3(g) + 2 NaI(aq) + H2O(l) → O2(g) + I2(s) + 2 NaOH(aq) How many moles of sodium iodide are needed to remove 4.54 ✕ 10−6 mol of O3? mol NaI How many milligrams of sodium iodide are needed to remove 13.31 mg of O3?

Answer 1

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by =$\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =$\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$    (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$.

Answer 2

$9.08 \times 10^-^6$ moles of sodium iodide are needed to remove 4.54 ✕ 10−6 mol of O3.

The mass of sodium iodide is 90 mg.

What are moles?

In the International System of Units, Mole is the base unit of the amount of any substance.

The reaction is

$\rm O_3(g) + 2 NaI(aq) + H2O(l) =O2(g) + I2(s) + 2 NaOH(aq)$

If 2 moles of sodium iodide are needed to remove 1 mole of ozone

Then, for $4.54 \times 10^-^6 mol\; of O_3$

$\dfrac{1}{2} \times 4.54 \times 10^-^6=9.08\times 10^-^6$

Number of moles of ozone is

$Number\;of \;moles= \dfrac{mass}{molar\;mass}\\\\\dfrac{0.01331g}{48/mol} =0.0003 moles$

Now, the same for 0.0003 moles

$\dfrac{1}{2} \times 0.0003 = 0.0006\;mol$

Calculate the mass

mass = molar mass x moles

$0.0003 \times 150\;g/mol = 0.09 g = 90\;mg$

Thus, the moles of sodium iodide are $9.08 \times 10^-^6$

The mass of sodium iodide is 90 mg.

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