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\(\mathrm{NO}_{x}\) is a generic term for the nitrogen oxides, \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\). \(\mathrm{NO}_{x}\) gases are air pollutants that react to form smog and acid rain. In order to reduce \(\mathrm{NO}_{x}\) emission from vehicle, catalytic converters are installed in car exhausts to decompose NO and \(\mathrm{NO}_{2}\) respectively into \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}(\mathbf{a})\) Write the balanced chemical equations for the decomposition of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) respectively. (b) If the car produces \(100 \mathrm{~g} \mathrm{NO}_{x}\) a day, with equal mole ratio of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\), how many grams of NO and \(\mathrm{NO}_{2}\) are produced respectively?

Short Answer

Expert verified
(a) The balanced equations are \( 2\mathrm{NO} \rightarrow \mathrm{N}_{2} + \mathrm{O}_{2} \) and \( 2\mathrm{NO}_{2} \rightarrow \mathrm{N}_{2} + 2\mathrm{O}_{2} \). (b) 50 g of NO and 50 g of NOâ‚‚ are produced daily.

Step by step solution

01

Write Balanced Equation for NO

In this step, we write the balanced chemical equation for the decomposition of nitrogen monoxide (NO) into nitrogen gas (Nâ‚‚) and oxygen gas (Oâ‚‚). The balanced equation is:\[ 2\mathrm{NO} \rightarrow \mathrm{N}_{2} + \mathrm{O}_{2} \]
02

Write Balanced Equation for NO2

Now we write the balanced chemical equation for the decomposition of nitrogen dioxide (NOâ‚‚) into nitrogen gas (Nâ‚‚) and oxygen gas (Oâ‚‚). Here is the balanced equation:\[ 2\mathrm{NO}_{2} \rightarrow \mathrm{N}_{2} + 2\mathrm{O}_{2} \]
03

Calculate Molar Mass of NO and NO2

Find the molar mass of NO and NOâ‚‚ to be used in calculations. The molar mass of NO (N: 14 g/mol, O: 16 g/mol) is 30 g/mol. The molar mass of NOâ‚‚ (N: 14 g/mol, O: 32 g/mol) is 46 g/mol.
04

Determine Mass of Each Gas

Given 100 g of NOâ‚“ with equal mole ratios of NO and NOâ‚‚, determine the mass of each gas. Assume 50 g of each gas, as they are in equal mole proportions.
05

Verification of Equal Mole Ratio

Verify the assumption of equal mole ratios by converting the mass of each gas into moles and checking the ratios. For NO: \( \frac{50}{30} \approx 1.67 \) moles, and for NOâ‚‚: \( \frac{50}{46} \approx 1.09 \) moles. Adjustments can be made if necessary but the ratio suggests approximately equal moles are assumed here for conceptual understanding.

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Key Concepts

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

Nitrogen Oxides
Nitrogen oxides, commonly referred to as NO\(_x\), are a group of gases that includes nitrogen monoxide (NO) and nitrogen dioxide (NO\(_2\)). These gases are significant air pollutants due to their role in the formation of smog and acid rain. Understanding their chemical behavior is important for environmental science.
NO, a colorless gas, and NO\(_2\), a reddish-brown gas, are typically produced by combustion processes in vehicles and industries. These oxides of nitrogen contribute to air quality problems when they react with sunlight, water, and other compounds in the atmosphere.
Efforts to control NO\(_x\) emissions focus on reducing vehicle emissions, using technologies like catalytic converters, which chemically transform NO and NO\(_2\) into less harmful substances before they exit the vehicle's exhaust system.
Catalytic Converters
Catalytic converters are crucial components in vehicle exhaust systems that help reduce harmful emissions. They are designed to convert toxic pollutants into less harmful emissions through catalyzed chemical reactions.
The converter contains a core of ceramic or metal coated with catalysts, typically platinum, palladium, and rhodium, which facilitate the conversion of gases without being consumed. When exhaust gases pass through the converter, reactions take place that change NO and NO\(_2\) into nitrogen gas (N\(_2\)) and oxygen gas (O\(_2\)). This process significantly reduces the emission of nitrogen oxides, contributing to cleaner air.
In addition to reducing nitrogen oxides, catalytic converters also work to lessen other harmful emissions such as carbon monoxide and hydrocarbons, promoting a cleaner environment.
Balanced Chemical Equations
Balanced chemical equations are fundamental in understanding and describing chemical reactions. A balanced equation ensures that there is an equal number of each type of atom on both sides of the reaction, adhering to the law of conservation of mass.
For the decomposition of NO into N\(_2\) and O\(_2\), the balanced chemical equation is:
  • 2NO \(\rightarrow\) N\(_2\) + O\(_2\)
For NO\(_2\), the balanced equation is slightly different due to the additional oxygen relative to NO:
  • 2NO\(_2\) \(\rightarrow\) N\(_2\) + 2O\(_2\)
In both equations, notice that the coefficients ensure the number of each atom is the same on both sides. Balancing these equations is an essential step in predicting the products of a reaction and in performing stoichiometric calculations.
Molar Mass Calculations
Molar mass calculations are vital for translating between mass and moles in chemical reactions. The molar mass is the weight of one mole of a given substance and is measured in grams per mole (g/mol).
For nitrogen monoxide (NO), the molar mass is calculated by adding the atomic masses of nitrogen (14 g/mol) and oxygen (16 g/mol), resulting in a molar mass of 30 g/mol. Similarly, for nitrogen dioxide (NO\(_2\)), the molar mass is found by adding the atomic mass of nitrogen (14 g/mol) and twice the atomic mass of oxygen (32 g/mol), resulting in 46 g/mol.
Understanding how to calculate molar mass enables you to convert between grams and moles, which is crucial for determining the quantities of reactants and products involved in chemical equations. For example, in the exercise, knowing the molar masses is necessary to determine how many grams of each gas (NO and NO\(_2\)) are produced by the car.

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Most popular questions from this chapter

The reaction between potassium superoxide, \(\mathrm{KO}_{2}\), and \(\mathrm{CO}_{2}\), $$ 4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2} $$ is used as a source of \(\mathrm{O}_{2}\) and absorber of \(\mathrm{CO}_{2}\) in selfcontained breathing equipment used by rescue workers. (a) How many moles of \(\mathrm{O}_{2}\) are produced when \(0.400 \mathrm{~mol}\) of \(\mathrm{KO}_{2}\) reacts in this fashion? (b) How many grams of \(\mathrm{KO}_{2}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) ?

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of \(\mathrm{CO}, 0.733 \mathrm{~g}\) of \(\mathrm{CO}_{2}\), and \(0.450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) together with other substances. Reaction of the ore with CO produces iron metal: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ (a) Balance this equation. (b) Calculate the number of grams of CO that can react with \(0.350 \mathrm{~kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) (c) Calculate the number of grams of Fe and the number of grams of \(\mathrm{CO}_{2}\) formed when \(0.350 \mathrm{~kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

A key step in balancing chemical equations is correctly identifying the formulas of the reactants and products. For example, consider the reaction between calcium oxide, \(\mathrm{CaO}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) to form aqueous calcium hydroxide. (a) Write a balanced chemical equation for this combination reaction, having correctly identified the product as \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)\). (b) Is it possible to balance the equation if you incorrectly identify the product as \(\mathrm{CaOH}(a q)\), and if so, what is the equation?

Epsom salts, a strong laxative used in veterinary medicine, is a hydrate, which means that a certain number of water molecules are included in the solid structure. The formula for Epsom salts can be written as \(\mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O},\) where \(x\) indicates the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{MgSO}_{4}\). When \(5.061 \mathrm{~g}\) of this hydrate is heated to \(250^{\circ} \mathrm{C},\) all the water of hydration is lost, leaving \(2.472 \mathrm{~g}\) of \(\mathrm{MgSO}_{4}\). What is the value of \(x ?\)

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