Question

The depletion of ozone \(\left(\mathrm{O}_{3}\right)\) in the stratosphere has been a matter of great concern among scientists in recent years. It is believed that ozone can react with nitric oxide (NO) that is discharged from high-altitude jet planes. The reaction is $$ \mathrm{O}_{3}+\mathrm{NO} \longrightarrow \mathrm{O}_{2}+\mathrm{NO}_{2} $$ If \(0.740 \mathrm{~g}\) of \(\mathrm{O}_{3}\) reacts with \(0.670 \mathrm{~g}\) of NO, how many grams of \(\mathrm{NO}_{2}\) will be produced? Which compound is the limiting reactant? Calculate the number of moles of the excess reactant remaining at the end of the reaction.

Short answer

0.709 g of NO₂ will be produced; O₃ is the limiting reactant; 0.0069 mol of NO remains.

Step-by-step solution

Determine the Molar Masses

Calculate the molar masses of \( \text{O}_3 \) and \( \text{NO} \) using the periodic table. For \( \text{O}_3 \), each oxygen atom has a molar mass of approximately 16.00 g/mol, so the molar mass of \( \text{O}_3 \) is \( 3 \times 16.00 = 48.00 \) g/mol. For \( \text{NO} \), the nitrogen atom has a molar mass of approximately 14.01 g/mol, and oxygen is 16.00 g/mol, so the molar mass of \( \text{NO} \) is \( 14.01 + 16.00 = 30.01 \) g/mol.

Calculate the Moles of Each Reactant

Use the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \) to find the moles of each reactant. For \( \text{O}_3 \), use \( \frac{0.740 \, g}{48.00 \, g/mol} \approx 0.0154 \, \text{mol} \). For \( \text{NO} \), use \( \frac{0.670 \, g}{30.01 \, g/mol} \approx 0.0223 \, \text{mol} \).

Identify the Limiting Reactant

Examine the stoichiometry of the reaction: 1 mole of \( \text{O}_3 \) reacts with 1 mole of \( \text{NO} \). Since we have 0.0154 mol of \( \text{O}_3 \) and 0.0223 mol of \( \text{NO} \), \( \text{O}_3 \) is the limiting reactant because there are fewer moles of it relative to the stoichiometric ratio.

Calculate Moles of Product Formed

Since \( \text{O}_3 \) is the limiting reactant, it determines the amount of product formed. The reaction is 1:1 to produce \( \text{NO}_2 \). Therefore, \( 0.0154 \) mol of \( \text{NO}_2 \) is produced because \( 0.0154 \) mol of \( \text{O}_3 \) is consumed.

Convert Moles of Product to Grams

Find the molar mass of \( \text{NO}_2 \), which is \( 14.01 \text{ (N)} + 2 \times 16.00 \text{ (O)} = 46.01 \, g/mol \). Multiply the moles of \( \text{NO}_2 \) by its molar mass: \( 0.0154 \, \text{mol} \times 46.01 \, g/mol \approx 0.709 \text{ g of NO}_2 \).

Calculate Remaining Moles of Excess Reactant

Since \( \text{NO} \) is in excess, calculate how much remains after the reaction: 0.0154 mol \( \text{NO} \) reacts with \( \, 0.0154\) mol \( \text{O}_3 \), leaving \( 0.0223 - 0.0154 \approx 0.0069 \, \text{mol} \) of \( \text{NO} \) remaining.

Key concepts

Stoichiometry

Stoichiometry is the method used by chemists to determine the quantities of reactants and products involved in a chemical reaction. It's like a recipe for chemical equations, showing the relationship between the amounts of substances. When dealing with stoichiometry, it is important to understand the balanced chemical equation, as it provides the exact ratios needed for the reaction to proceed. In this exercise, the balanced equation is:

• \[ \mathrm{O}_3 + \mathrm{NO} \rightarrow \mathrm{O}_2 + \mathrm{NO}_2\]

This equation tells us that one mole of ozone reacts with one mole of nitric oxide to produce one mole of oxygen and one mole of nitrogen dioxide. By understanding this ratio (1:1:1:1), we can calculate other components, such as identifying the limiting reactant, which restricts the amount of product formed.
Analyzing the stoichiometry helps in defining which reactant will run out first and consequently halt the reaction. Knowing how to manipulate these relationships is crucial for solving chemical equation problems.

Molar Mass

Molar mass is critical for converting between grams and moles of a substance, acting as a bridge for calculations in chemistry. It is the mass of one mole of a given substance, usually expressed in grams per mole (g/mol). The molar mass is determined using the atomic masses of the elements found on the periodic table. In this problem, we needed the molar masses of all reactants and one product:

• For \( \mathrm{O}_3 \), the molar mass is calculated by multiplying the atomic mass of oxygen (approximately 16.00 g/mol) by three, giving us \( 48.00 \) g/mol.


• For \( \mathrm{NO} \), combining nitrogen’s atomic mass (14.01 g/mol) and oxygen’s, we get \( 30.01 \) g/mol.


• For the product \( \mathrm{NO}_2 \), the molar mass becomes \( \text{14.01 (N)} + 2 \times 16.00 \text{ (O)} = 46.01 \, g/mol \).

By understanding molar mass, one can easily convert from the mass of a compound to the amount in moles, which is a fundamental step in any stoichiometric calculation.

Moles Calculation

Calculating moles from given mass involves applying the formula:

• \[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

This is vital for relating the mass of a substance to the number of particles, as reactions occur on a molecular level, not by the mass we measure on a scale.
In this problem:

• For \( \mathrm{O}_3 \), given \( 0.740 \) g, moles are calculated as:\[\frac{0.740 \, \text{g}}{48.00 \, \text{g/mol}} \approx 0.0154 \, \text{mol} \]


• For \( \mathrm{NO} \), given \( 0.670 \) g, it becomes:\[\frac{0.670 \, \text{g}}{30.01 \, \text{g/mol}} \approx 0.0223 \, \text{mol}\]

These calculations guide you in determining the limiting reactant, allowing the calculation of any remaining reactants and products.

Chemical Reactions

Understanding chemical reactions involves looking at how substances interact to form new products. Every chemical reaction can be depicted by a balanced chemical equation that outlines how reactants turn into products under specific conditions. For reactions like the one in the exercise:Ozone reacts with nitric oxide to produce oxygen and nitrogen dioxide. This is a type of replacement reaction where molecules exchange partners. Knowing the nature of the reactants and products involved helps predict the outcomes and any possible remaining substances.
It's important to know the initial amounts of reactants and how they determine the path of the reaction. When one reactant is used up (like \( \mathrm{O}_3 \) here, the limiting reactant), the reaction stops, and you might have some of another reactant left over (in this case, \( \mathrm{NO} \) was calculated to be in excess).
Recognizing types of chemical reactions aids in determining both the potential yield of a reaction and any practical implications. These insights are vital to both controlled laboratory settings and real-world scenarios like environmental chemistry.