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Chapter 3: Problem 97

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 \(\mathrm{O}_{3} ?\)

Short Answer

Expert verified
(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.

Step by step solution

01

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.
02

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:
03

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.
04

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.
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.
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.
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.

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Most popular questions from this chapter

Vanillin, the dominant flavoring in vanilla, contains C, H, and O. When \(1.05 \mathrm{~g}\) of this substance is completely combusted, \(2.43 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.50 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced. What is the empirical formula of vanillin?

(a) What is the mass, in grams, of \(1.223 \mathrm{~mol}\) of iron(III) sulfate? (b) How many moles of ammonium ions are in \(6.955 \mathrm{~g}\) of ammonium carbonate? (c) What is the mass, in grams, of \(1.50 \times 10^{21}\) molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4} ?\) (d) What is the molar mass of diazepam (Valium \(^{\circ}\) ) if 0.05570 mol has a mass of \(15.86 \mathrm{~g}\) ?

(a) If an automobile travels \(225 \mathrm{mi}\) with a gas mileage of \(20.5 \mathrm{mi} / \mathrm{gal}\), how many kilograms of \(\mathrm{CO}_{2}\) are produced? Assume that the gasoline is composed of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l),\) whose density is \(0.69 \mathrm{~g} / \mathrm{mL}\). (b) Repeat the calculation for a truck that has a gas mileage of \(5 \mathrm{mi} / \mathrm{gal}\).

(a) When the metallic element sodium combines with the nonmetallic element bromine, \(\mathrm{Br}_{2}(l),\) how can you determine the chemical formula of the product? How do you know whether the product is a solid, liquid, or gas at room temperature? Write the balanced chemical equation for the reaction. (b) When a hydrocarbon burns in air, what reactant besides the hydrocarbon is involved in the reaction? What products are formed? Write a balanced chemical equation for the combustion of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l),\) in air.

Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{PbCO}_{3}(s) \longrightarrow \mathrm{PbO}(s)+\mathrm{CO}_{2}(g)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{Mg}(s)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Mg}_{3} \mathrm{~N}_{2}(s)\) (d) \(\mathrm{C}_{7} \mathrm{H}_{8} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (e) \(\mathrm{Al}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{AlCl}_{3}(s)\)
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