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Chapter 9: Problem 51

Nitrogen monoxide, NO, reacts with oxygen to form nitrogen dioxide. Then the nitrogen dioxide reacts with oxygen to form nitrogen monoxide and ozone. Write the balanced equations. What is the theoretical yield in grams of ozone from 4.55 g of nitrogen monoxide with excess \(\mathrm{O}_{2} ?\) (Hint: First calcu- late the theoretical yield for NO \(_{2}\) , then use that value to calculate the yield for ozone.)

Short Answer

Expert verified
The theoretical yield of ozone from 4.55g of nitrogen monoxide with excess \(O_{2}\) is approximately 3.64g.

Step by step solution

01

Balancing the Chemical Reactions

The two reactions are: \[ \mathrm{2NO + O_{2} \rightarrow 2NO_{2}} \] and \[ \mathrm{2NO_{2} + O_{2} \rightarrow 2NO + O_{3}} \].
02

Calculate Theoretical Yield for NO2

First convert the amount of NO to moles using its molar mass \(30.01 \, g/mol\): \[ \mathrm{Moles\, of\, NO = \frac{4.55\, g}{30.01\, g/mol} = 0.1516\, mol} \]. From the balanced equation, the mole ratio of NO to NO2 is 1:1, so the moles of NO2 produced is also \(0.1516\, mol\).
03

Calculate Theoretical Yield for Ozone

Next, use the molar ratio from the second equation to find the moles of O3 produced from NO2: \[ \mathrm{Moles\, of\, O_{3} = \frac{0.1516\, mol\, of\, NO_{2}}{2} = 0.0758\, mol} \]. Multiply by the molar mass of O3, \(48.00 \, g/mol\), to obtain the mass in grams: \[ \mathrm{Theoretical\, Yield\, of\, O_{3} = 0.0758\, mol * 48.00\, g/mol = 3.6384\, g} \]. Ozone production is limited by the amount of NO present, not O2, therefore the theoretical yield is based on the initial amount of NO.

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Most popular questions from this chapter

Use the balanced equation below to write mole ratios for the situations that follow. $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)$$ \begin{equation}\begin{array}{l}{\text { a. calculating mol } \mathrm{H}_{2} \mathrm{O} \text { given mol } \mathrm{H}_{2}} \\ {\text { b. calculating mol } \mathrm{O}_{2} \text { given mol } \mathrm{H}_{2} \mathrm{O}} \\ {\text { c. calculating mol } \mathrm{H}_{2} \text { given mol } \mathrm{O}_{2}}\end{array}\end{equation}

How many grams of Al form from 9.73 g of aluminum oxide if the yield is 91\(\% ?\) $$\mathrm{Al}_{2} \mathrm{O}_{3}+3 \mathrm{C} \rightarrow 2 \mathrm{Al}+3 \mathrm{CO}$$

Why is the ratio of fuel to air in a car's engine important in controlling pollution?

Use the equation provided to answer the questions that follow. $$2 \mathrm{NO}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}$$ \begin{equation}\begin{array}{l}{\text { a. How many molecules of } \mathrm{NO}_{2} \text { can form }} \\ {\text { from } 1.11 \mathrm{mol} \mathrm{O}_{2} \text { and excess } \mathrm{NO} \text { ? }} \\ {\text { b. How many molecules of NO will react }} \\ {\text { with } 25.7 \mathrm{g} \mathrm{O}_{2} ?} \\ {\text { c. How many molecules of } \mathrm{O}_{2} \text { are needed to }} \\ {\text { make } 3.76 \times 10^{22} \text { molecules } \mathrm{NO}_{2} ?}\end{array}\end{equation}

In the reaction shown below, 64 \(\mathrm{g} \mathrm{CaC}_{2}\) is reacted with 64 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) $$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s)$$ \begin{equation}\begin{array}{l}{\text { a. Which is the excess reactant, and which is }} \\ {\text { the limiting reactant? }} \\ {\text { b. What is the theoretical yield of } C_{2} \mathrm{H}_{2} ?} \\ {\text { c. What is the theoretical yield of } \mathrm{Ca}(\mathrm{OH})_{2} ?}\end{array}\end{equation}
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