Problem 97
Question
A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the air sample through a "bubbler" containing sodium iodide, which removes the ozone according to the following equation: \(\mathrm{O}_{3}(g)+2 \mathrm{NaI}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$ \mathrm{O}_{2}(g)+\mathrm{I}_{2}(s)+2 \mathrm{NaOH}(a q) $$ (a) How many moles of sodium iodide are needed to remove \(5.95 \times 10^{-6} \mathrm{~mol}\) of \(\mathrm{O}_{3} ?\) (b) How many grams of sodium iodide are needed to remove \(1.3 \mathrm{mg}\) of \(\mathrm{O}_{3} ?\)
Step-by-Step Solution
Verified Answer
(a) \(1.19 \times 10^{-5}\) mol of NaI are needed to remove \(5.95 \times 10^{-6}\) mol of O3.
(b) 8.12 mg of NaI are needed to remove 1.3 mg of O3.
1Step 1: Find the mole ratio of NaI to O3 in the balanced equation
The given balanced chemical equation is: O3(g) + 2 NaI(aq) + H2O(l) → O2(g) + I2(s) + 2 NaOH(aq). From this equation, we can see that 2 moles of NaI are needed to remove 1 mole of O3.
2Step 2: Calculate the moles of NaI needed
Since 2 moles of NaI react with 1 mole of O3, we can use the mole ratio to find the amount of NaI needed for 5.95x10⁻⁶ mol of O3: moles of NaI = (5.95x10⁻⁶ mol O3) x (2 mol NaI / 1 mol O3) = 1.19x10⁻⁵ mol NaI.
(b) Grams of NaI needed for 1.3 mg of O3:
3Step 3: Convert mass of O3 to moles
First, convert the mass of O3 from mg to grams: 1.3 mg = 1.3x10⁻³ g. Next, find the molar mass of O3: Molar mass of O3 = 3 x 16.00 g/mol (Oxygen has a molar mass of 16.00 g/mol) = 48.00 g/mol. Then, convert the grams of O3 to moles: moles of O3 = 1.3x10⁻³ g ÷ 48.00 g/mol ≈ 2.708x10⁻⁵ mol O3.
4Step 4: Calculate the moles of NaI needed
Repeat the calculation from step 2 to find the moles of NaI needed for 2.708x10⁻⁵ mol of O3: moles of NaI = (2.708x10⁻⁵ mol O3) x (2 mol NaI / 1 mol O3) ≈ 5.416x10⁻⁵ mol NaI.
5Step 5: Convert moles of NaI to grams
Find the molar mass of NaI: Molar mass of NaI = 22.99 g/mol (Na) + 126.9 g/mol (I) = 149.89 g/mol. Then, convert the moles of NaI to grams: grams of NaI = (5.416x10⁻⁵ mol NaI) x (149.89 g/mol) ≈ 8.12x10⁻³ g ≈ 8.12 mg NaI.
So, (a) 1.19x10⁻⁵ mol of NaI are needed to remove 5.95x10⁻⁶ mol of O3, and (b) 8.12 mg of NaI are needed to remove 1.3 mg of O3.
Key Concepts
Understanding Mole RatioDecoding Chemical ReactionsNavigating Molar Mass Calculations
Understanding Mole Ratio
The mole ratio in a chemical reaction is a crucial element for understanding how reactants combine to form products. In the context of our example, the mole ratio is used to determine the proportion of sodium iodide (NaI) that reacts with ozone (O3). Based on the balanced chemical equation, we see a specific pattern: for every one mole of O3, two moles of NaI are required.
This 1:2 ratio can be used as a conversion factor to calculate the exact amount of NaI needed to react with a given amount of ozone. For instance, to remove 5.95x10−6 moles of O3, we multiply by the mole ratio (2 moles NaI / 1 mole O3) to get 1.19x10−5 moles of NaI. This clearly illustrates how mole ratios serve as the bridge between the quantities of different chemicals in reactions.
This 1:2 ratio can be used as a conversion factor to calculate the exact amount of NaI needed to react with a given amount of ozone. For instance, to remove 5.95x10−6 moles of O3, we multiply by the mole ratio (2 moles NaI / 1 mole O3) to get 1.19x10−5 moles of NaI. This clearly illustrates how mole ratios serve as the bridge between the quantities of different chemicals in reactions.
Decoding Chemical Reactions
Chemical reactions are the heart of chemistry, where substances transform into new products. Every reaction is governed by the Law of Conservation of Mass, meaning matter cannot be created or destroyed. Therefore, a balanced chemical equation is essential as it shows a reaction where each atom's count remains constant.
When we examine the ozone-removal reaction, we notice that for every molecule of O3 that reacts, an entire set of products is formed, including oxygen gas (O2), solid iodine (I2), and aqueous sodium hydroxide (NaOH). Understanding the stoichiometry of the reaction allows us to predict the outcomes, whether it's the amounts of products formed or the reactants consumed.
When we examine the ozone-removal reaction, we notice that for every molecule of O3 that reacts, an entire set of products is formed, including oxygen gas (O2), solid iodine (I2), and aqueous sodium hydroxide (NaOH). Understanding the stoichiometry of the reaction allows us to predict the outcomes, whether it's the amounts of products formed or the reactants consumed.
Navigating Molar Mass Calculations
Molar mass calculations are integral to converting between the weight of a substance and the moles of a substance. Each element has a unique molar mass, typically expressed in grams per mole (g/mol), which corresponds to its relative atomic mass. To find the molar mass of a compound like NaI, we sum the molar masses of sodium (Na) and iodine (I).
Once we have the molar mass, we can convert from grams to moles or vice versa using the equation:
moles = mass in grams / molar mass (g/mol).
From our example, converting 1.3 mg of O3 to moles and then to grams of NaI relies on these calculations. Accurate molar mass determinations are imperative to ensure precise stoichiometric calculations in chemistry.
Once we have the molar mass, we can convert from grams to moles or vice versa using the equation:
moles = mass in grams / molar mass (g/mol).
From our example, converting 1.3 mg of O3 to moles and then to grams of NaI relies on these calculations. Accurate molar mass determinations are imperative to ensure precise stoichiometric calculations in chemistry.
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