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Chapter 9: Problem 45

Nitrogen dioxide from exhaust reacts with oxygen to form ozone. What mass of ozone could be formed from 4.55 \(\mathrm{g} \mathrm{NO}_{2} ?\) If only 4.58 \(\mathrm{g} \mathrm{O}_{3}\) formed, what is the percentage yield? $$\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{O}_{3}(g)$$

Short Answer

Expert verified
The theoretical yield of \(O_3\) is 4.8 g and the actual yield is 4.58 g. The percentage yield of the reaction is 95.4%.

Step by step solution

01

Determine moles of \(NO_2\)

We start by converting the given mass of \(NO_2\) to moles, using the molar mass of \(NO_2\) (46.01 g/mol). \n\[ Moles\:of\:NO_2 = \frac{4.55\:g}{46.01\:g/mol} = 0.1\:mol. \]
02

Calculate theoretical yield of \(O_3\)

The balanced equation shows us that one mole of \(NO_2\) forms one mole of \(O_3\). This means the stoichiometric ratio of \(NO_2\) to \(O_3\) is 1:1. So, we should have theoretically produced 0.1 moles of \(O_3\). We'll then convert this to grams using the molar mass of \(O_3\) (48.00 g/mol).\n\[Theoretical\:yield\:of\:O_3 = Molar\:mass\:of\:O_3 \times Moles\:of\:O_3 = 0.1\:mol \times 48.00\:g/mol = 4.8\:g. \]
03

Calculate percentage yield

The given mass of \(O_3\) formed in the experiment is 4.58 g, which is less than the theoretical yield of 4.8 g. The percentage yield is calculated as follows:\n\[Percentage\:yield = \frac{Actual\:Yield}{Theoretical\:Yield} \times 100% = \frac{4.58\:g}{4.8\:g} \times 100% = 95.4\%. \] This means that 95.4% of the maximum possible amount of \(O_3\) was generated in the reaction.

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Most popular questions from this chapter

Describe the relationship between the limiting reactant and the theoretical yield.

Use the following terms to create a concept map: stoichiometry, excess reactant, theoretical yield, and mole ratio.

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.)

Use the equation provided to answer the questions that follow. The density of oxygen gas is 1.428 \(\mathrm{g} / \mathrm{L}\) . $$2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$$ \begin{equation}\begin{array}{l}{\text { a. What volume of oxygen can be made }} \\ {\text { from } 5.00 \times 10^{-2} \text { mol of } \mathrm{KClO}_{3} ?} \\ {\text { b. How many grams } \mathrm{KClO}_{3} \text { must react to }} \\\ {\text { form } 42.0 \mathrm{mL} \mathrm{O}_{2} \text { ? }} \\ {\text { c. How many milliliters of } \mathrm{O}_{2} \text { will form at }} \\ {\text { STP from } 55.2 \mathrm{g} \mathrm{KClO}_{3} ?}\end{array}\end{equation}

The percentage yield of nitric acid is 95\(\% .\) If 9.88 \(\mathrm{kg}\) of nitrogen dioxide react, what mass of nitric acid is isolated? $$3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$$
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