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Chapter 4: Problem 79

How many grams of commercial acetic acid (97\% \(\mathrm{CH}_{3} \mathrm{COOH}\) by mass) must be allowed to react with an excess of \(\mathrm{PCl}_{3}\) to produce \(75 \mathrm{g}\) of acetyl chloride \(\left(\mathrm{CH}_{3} \mathrm{COCl}\right),\) if the reaction has a \(78.2 \%\) yield? $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{PCl}_{3} & \longrightarrow \mathrm{CH}_{3} \mathrm{COCl}+ \mathrm{H}_{3} \mathrm{PO}_{3}(\text { not balanced }) \end{aligned}$$

Short Answer

Expert verified
75.5 grams of commercial acetic acid is needed.

Step by step solution

01

Balance the Chemical Equation

The balanced equation is:\[ \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{PCl}_{3} \longrightarrow \mathrm{CH}_{3} \mathrm{COCl}+ \mathrm{H}_{3} \mathrm{PO}_{3} \]
02

Calculate the Moles of Acetyl Chloride Produced

75g of acetyl chloride is produced. The molar mass of CH3COCl is \(3(12.01) + 1(1.01) + 16.00 + 35.45 = 78.5 \) g/mol, so divide the produced mass by molar mass:\[ \frac{75 g}{78.5 g/mol} = 0.955 moles \]
03

Calculate the Theoretical Yield

The reaction has a 78.2% yield. To find the theoretical yield, divide the number of moles of CH3COCl produced by the actual percent yield:\[ \frac{0.955 moles}{0.782} = 1.221 moles \]
04

Determine the moles of Acetic Acid

Because the equation is balanced, one mole acetic acid produces one mole acetyl chloride. Therefore, 1.221 moles of acetic acid were reacted.
05

Determine Grams of Commercial Acetic Acid

The molar mass of CH3COOH is \(3(12.01) + 1(1.01) + 16.00 + 16.00 + 1(1.01) = 60.05 \) g/mol. Converting moles of acetic acid to grams:\[ 1.221 moles * 60.05 g/mol = 73.3 g \]Given that it's 97% acetic acid by mass, we can determine how many grams of the commercial acetic acid is needed:\[ \frac{73.3 g}{0.97} = 75.5 g \]

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Most popular questions from this chapter

A method of lowering the concentration of \(\mathrm{HCl}(\mathrm{aq})\) is to allow the solution to react with a small quantity of Mg. How many milligrams of Mg must be added to \(250.0 \mathrm{mL}\) of \(1.023 \mathrm{M} \mathrm{HCl}\) to reduce the solution concentration to exactly \(1.000 \mathrm{M} \mathrm{HCl} ?\) $$ \mathrm{Mg}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) $$

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When the equation below is balanced, the correct set of stoichiometric coefficients is (a) \(1,6 \longrightarrow 1,3,4;\) (b) \(1,4 \longrightarrow 1,2,2 ;\) (c) \(2,6 \longrightarrow 2,3,2;\) (d) \(3,8 \longrightarrow 3,4,2\) \(\begin{aligned} ? \mathrm{Cu}(\mathrm{s})+? \mathrm{HNO}_{3}(\mathrm{aq}) & \longrightarrow ? \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+? \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+? \mathrm{NO}(\mathrm{g}) \end{aligned}\)

For a specific reaction, ammonium dichromate is the only reactant and chromium(III) oxide and water are two of the three products. What is the third product and how many grams of this product are produced per kilogram of ammonium dichromate decomposed?

Consider the chemical equation below. What is the maximum number of moles of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) that can be obtained from a reaction mixture containing 5.0 moles each of \(\mathrm{KMnO}_{4}, \mathrm{KI},\) and \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\) (a) \(3.0 \mathrm{mol}\); (b) \(3.8 \mathrm{mol}\) (c) \(5.0 \mathrm{mol} ;\) (d) \(6.0 \mathrm{mol} ;\) (e) \(15 \mathrm{mol}\). $$\begin{array}{r} 2 \mathrm{KMnO}_{4}+10 \mathrm{KI}+8 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 6 \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+5 \mathrm{I}_{2}+8 \mathrm{H}_{2} \mathrm{O} \end{array}$$
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