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Chapter 4: Problem 74

In the reaction shown, \(100.0 \mathrm{g} \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) yielded \(64.0 \mathrm{g}\) \(\mathrm{C}_{6} \mathrm{H}_{10} .\) (a) What is the theoretical yield of the reaction? (b) What is the percent yield? (c) What mass of \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) would produce \(100.0 \mathrm{g} \mathrm{C}_{6} \mathrm{H}_{10}\) if the percent yield is that determined in part (b)? $$\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{10}+\mathrm{H}_{2} \mathrm{O}$$

Short Answer

Expert verified
The theoretical yield of the reaction is approximately 95.4g. The actual yield being 64.0g yields a percent yield of roughly 67.0%. This means that to produce 100.0g of \(\mathrm{C}_{6}\mathrm{H}_{10}\), it would require about 144.7g of \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\).

Step by step solution

01

Calculate Theoretical Yield

To calculate the theoretical yield, find the number of moles of the reactant \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) by using its molar mass (86.15 g/mol). Then, relate this to the moles of the product \(\mathrm{C}_{6}\mathrm{H}_{10}\) using the stoichiometric ratio of 1 to 1 given by the balanced equation. Further, convert the moles of the product back to grams using its molar mass (82.15 g/mol). The formula would look like: \[ \text{No. of moles of } \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}= \frac{100.0g}{86.15 g/mol} \] \[ \text{Theoretical Yield of } \mathrm{C}_{6}\mathrm{H}_{10}=\text{No. of moles of } \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH} \times 82.15 g/mol \]
02

Calculate Percent Yield

To calculate the percent yield, divide the actual yield by the theoretical yield and multiply the quotient by 100%. The percent yield formula would look like: \[ \text{Percent yield} = \frac{64.0g}{\text{Theoretical Yield}} \times 100\% \]
03

Determine Initial Mass

To calculate the initial mass of the reactant that will yield 100.0g of the product given the calculated percent yield, first, figure out the theoretical yield of reactant needed to produce 100.0g of the product. Then, divide by the percent yield to account for the reaction's efficiency. The formula would look like: \[ \text{Theoretical yield of } \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH} = \frac{100.0g}{82.15 g/mol} \] \[ \text{Initial mass of } \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH} = \frac{\text{Theoretical yield of } \mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}}{\text{Percent yield}} \times 100\% \]

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Most popular questions from this chapter

Explain the important distinctions between (a) chemical formula and chemical equation; (b) stoichiometric coefficient and stoichiometric factor; (c) solute and solvent; (d) actual yield and percent yield; (e) consecutive and simultaneous reactions.

The following set of reactions is to be used as the basis of a method for producing nitric acid, \(\mathrm{HNO}_{3}\) Calculate the minimum masses of \(\mathrm{N}_{2}, \mathrm{H}_{2^{\prime}}\) and \(\mathrm{O}_{2}\) required per kilogram of \(\mathrm{HNO}_{3}\) $$\begin{array}{l} \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) \\ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) \\ 3 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) \end{array}$$

Nitrogen gas, \(\mathrm{N}_{2}\), can be prepared by passing gaseous ammonia over solid copper(II) oxide, \(\mathrm{CuO}\), at high temperatures. The other products of the reaction are solid copper, \(\mathrm{Cu},\) and water vapor. In a certain experiment, a reaction mixture containing \(18.1 \mathrm{g} \mathrm{NH}_{3}\) and \(90.4 \mathrm{g}\) CuO yields \(6.63 \mathrm{g} \mathrm{N}_{2}\). Calculate the percent yield for this experiment.

Titanium tetrachloride, \(\mathrm{TiCl}_{4}\) is prepared by the reaction below. $$\begin{aligned} &3 \mathrm{TiO}_{2}(\mathrm{s})+4 \mathrm{C}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{TiCl}_{4}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{CO}(\mathrm{g}) \end{aligned}$$ What is the maximum mass of \(\mathrm{TiCl}_{4}\) that can be obtained from \(35 \mathrm{g} \mathrm{TiO}_{2^{\prime}} 45 \mathrm{g} \mathrm{Cl}_{2^{\prime}}\) and \(11 \mathrm{g} \mathrm{C} ?\)

Which has the higher concentration of sucrose: a \(46 \%\) sucrose solution by mass \((d=1.21 \mathrm{g} / \mathrm{mL}),\) or \(1.50 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) ? Explain your reasoning.
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