Problem 79
Question
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? $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{PCl}_{3} & \longrightarrow \mathrm{CH}_{3} \mathrm{COCl}+ \mathrm{H}_{3} \mathrm{PO}_{3}(\text { not balanced }) \end{aligned}$$
Step-by-Step Solution
Verified Answer
75.5 grams of commercial acetic acid is needed.
1Step 1: Balance the Chemical Equation
The balanced equation is:\[ \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{PCl}_{3} \longrightarrow \mathrm{CH}_{3} \mathrm{COCl}+ \mathrm{H}_{3} \mathrm{PO}_{3} \]
2Step 2: 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:\[ \frac{75 g}{78.5 g/mol} = 0.955 moles \]
3Step 3: 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:\[ \frac{0.955 moles}{0.782} = 1.221 moles \]
4Step 4: 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.
5Step 5: 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:\[ \frac{73.3 g}{0.97} = 75.5 g \]
Key Concepts
Chemical ReactionsPercent YieldMolar MassBalanced Equation
Chemical Reactions
Chemical reactions are processes where substances (reactants) change to form new substances (products). This transformation involves breaking bonds in reactants and forming new bonds in products.
In the given reaction, acetic acid (\(\mathrm{CH}_{3} \mathrm{COOH}\)) reacts with phosphorus trichloride (\(\mathrm{PCl}_{3}\)) to produce acetyl chloride (\(\mathrm{CH}_{3} \mathrm{COCl}\)) and phosphonic acid (\(\mathrm{H}_{3} \mathrm{PO}_{3}\)).
Key points to remember about chemical reactions:
In the given reaction, acetic acid (\(\mathrm{CH}_{3} \mathrm{COOH}\)) reacts with phosphorus trichloride (\(\mathrm{PCl}_{3}\)) to produce acetyl chloride (\(\mathrm{CH}_{3} \mathrm{COCl}\)) and phosphonic acid (\(\mathrm{H}_{3} \mathrm{PO}_{3}\)).
Key points to remember about chemical reactions:
- Reactants and products have different properties.
- Energy is either absorbed or released, depending on the reaction.
- A chemical equation summarizes the transformation of reactants to products.
Percent Yield
Percent yield is a measure of how effective a chemical reaction is. It is calculated by comparing the actual yield (what you get from the reaction) to the theoretical yield (what you expect to get based on stoichiometry).
The formula for percent yield is: \[\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\]This is a handy tool for evaluating real-world chemical reactions, which often don't reach their theoretical yield due to impurities, incomplete reactions, or other practical issues.
In this exercise, the reaction of acetic acid with \(\mathrm{PCl}_{3}\) is given a yield of 78.2%, meaning that only 78.2% of the expected amount of acetyl chloride is actually produced.
The formula for percent yield is: \[\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\]This is a handy tool for evaluating real-world chemical reactions, which often don't reach their theoretical yield due to impurities, incomplete reactions, or other practical issues.
In this exercise, the reaction of acetic acid with \(\mathrm{PCl}_{3}\) is given a yield of 78.2%, meaning that only 78.2% of the expected amount of acetyl chloride is actually produced.
Molar Mass
Molar mass is the mass of one mole of a substance and is typically expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a molecule's chemical formula.
For example, the molar mass of acetic acid, \(\mathrm{CH}_{3}\mathrm{COOH}\), is determined by adding the atomic masses:
Knowing the molar mass of each compound allows you to convert between grams and moles, which is crucial in stoichiometry. This conversion helps predict how much of a substance you will need or produce in a reaction.
For example, the molar mass of acetic acid, \(\mathrm{CH}_{3}\mathrm{COOH}\), is determined by adding the atomic masses:
- Carbon (C): 12.01 g/mol (3 atoms)
- Hydrogen (H): 1.01 g/mol (4 atoms)
- Oxygen (O): 16.00 g/mol (2 atoms)
Knowing the molar mass of each compound allows you to convert between grams and moles, which is crucial in stoichiometry. This conversion helps predict how much of a substance you will need or produce in a reaction.
Balanced Equation
A balanced equation is essential in stoichiometry as it represents the conservation of mass. Each side of the equation must have the same number of each type of atom.
For the reaction involving acetic acid:\[\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{PCl}_{3} \longrightarrow \mathrm{CH}_{3}\mathrm{COCl} + \mathrm{H}_{3}\mathrm{PO}_{3}\]This equation is already balanced, indicating a direct one-to-one correspondence for reactants and products: one molecule of \(\mathrm{CH}_{3}\mathrm{COOH}\), reacts with \(\mathrm{PCl}_{3}\) to produce one molecule each of \(\mathrm{CH}_{3}\mathrm{COCl}\) and \(\mathrm{H}_{3}\mathrm{PO}_{3}\).
Balancing equations helps us understand the exact proportions in which reactants combine and products form, which is key to calculating yields and the quantities of substances needed in a reaction.
For the reaction involving acetic acid:\[\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{PCl}_{3} \longrightarrow \mathrm{CH}_{3}\mathrm{COCl} + \mathrm{H}_{3}\mathrm{PO}_{3}\]This equation is already balanced, indicating a direct one-to-one correspondence for reactants and products: one molecule of \(\mathrm{CH}_{3}\mathrm{COOH}\), reacts with \(\mathrm{PCl}_{3}\) to produce one molecule each of \(\mathrm{CH}_{3}\mathrm{COCl}\) and \(\mathrm{H}_{3}\mathrm{PO}_{3}\).
Balancing equations helps us understand the exact proportions in which reactants combine and products form, which is key to calculating yields and the quantities of substances needed in a reaction.
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