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 $O_3$.