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Chapter 9: Problem 46

How many grams \(\mathrm{CO}_{2}\) form from the complete combustion of \(1.00 \mathrm{L} \mathrm{C}_{8} \mathrm{H}_{18},\) density 0.700 \(\mathrm{g} / \mathrm{mL} ?\) If only \(1.90 \times 10^{3} \mathrm{g} \mathrm{CO}_{2}\) form, what is the percentage yield?

Short Answer

Expert verified
The complete combustion of 1.00L \(\mathrm{C}_{8}\mathrm{H}_{18}\) should produce 2156g of \(\mathrm{CO}_{2}\). However, if only 1.90 x \(10^{3}\)g \(\mathrm{CO}_{2}\) is produced, the percentage yield is 88.1%.

Step by step solution

01

Calculate Moles of Octane

The first step is to calculate the number of moles of \(\mathrm{C}_{8} \mathrm{H}_{18}\). Given its volume and density, the mass can first be found, then it can be converted to moles using the molecular weight. \( Mass = Volume \times Density = 1.00L \times 0.700g/mL = 700g \) (Note that 1L = 1000mL). Then, using the molecular mass of \(\mathrm{C}_{8} \mathrm{H}_{18}\) (114.22g/mol), the number of moles can be calculated as \( Moles = Mass / Molar Mass = 700g / 114.22g/mol = 6.13mol \)
02

Determine Moles of Carbon Dioxide

Then, one needs to determine how many moles of \(\mathrm{CO}_{2}\) can theoretically form from the complete combustion of \(\mathrm{C}_{8} \mathrm{H}_{18}\). According to the chemical equation for combustion, 1mol of \(\mathrm{C}_{8} \mathrm{H}_{18}\) gives 8mol of \(\mathrm{CO}_{2}\). Therefore, \( Moles\;of\;\mathrm{CO}_{2} = 6.13mol\;\mathrm{C}_{8}\mathrm{H}_{18} \times \frac{8mol\;\mathrm{CO}_{2}}{1mol\;\mathrm{C}_{8}\mathrm{H}_{18}} = 49.0mol \)
03

Calculate Mass of Theoretical Carbon Dioxide

Next, convert this to the theoretical mass of \(\mathrm{CO}_{2}\) using the molecular weight of \(\mathrm{CO}_{2}\) (44.01g/mol), yielding \( Mass\;of\;\mathrm{CO}_{2} = 49.0mol \times 44.01g/mol= 2156g \)
04

Calculate Percentage Yield

Lastly, the percentage yield can be calculated relative to the actual yield of \(\mathrm{CO}_{2}\) (1.90 x \(10^{3}\)g). \( Percentage\;Yield = \frac{Actual\;Yield}{Theoretical\; Yield} \times 100\% = \frac{1.90 \times 10^{3}\;g}{2156g} \times 100\% = 88.1\% \)

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Most popular questions from this chapter

What conversion factor is used to convert from volume of a gas directly to moles at STP?

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}

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}

Use the equation provided to answer the questions that follow. $$2 \mathrm{Na}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+\mathrm{H}_{2}$$ \begin{equation}\begin{array}{l}{\text { a. How many molecules of } \mathrm{H}_{2} \text { could be }} \\ {\text { made from } 27.6 \mathrm{g} \mathrm{H}_{2} \mathrm{O} \text { ? }} \\ {\text { b. How many atoms of Na will completely }} \\ {\text { react with } 12.9 \mathrm{g} \mathrm{H}_{2} \mathrm{O} \text { ? }} \\ {\text { c. How many molecules of } \mathrm{H}_{2} \text { could form }} \\ {\text { when } 6.59 \times 10^{20} \text { atoms Na react? }}\end{array}\end{equation}

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? $$\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{O}_{3}(g)$$
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