The student has started from $9.3 \text{ g}$ of aniline. However, the student has managed to obtain $11 \text{ g}$ of dry acetanilide.
The % yield of this reaction is :-
Calculating Acetanilide Percent Yield
This solution explains the calculation for the percent yield of acetanilide prepared from aniline and acetic anhydride, using the provided experimental masses.
Reaction Stoichiometry
The synthesis of acetanilide from aniline and acetic anhydride follows this reaction:
$ \text{C}_6\text{H}_5\text{NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{NHCOCH}_3 + \text{CH}_3\text{COOH} $
The molar ratio between aniline ($\text{C}_6\text{H}_5\text{NH}_2$) and acetanilide ($\text{C}_6\text{H}_5\text{NHCOCH}_3$) is 1:1.
Step 1: Calculate Moles of Aniline
Approximate molar mass of aniline ($\text{C}_6\text{H}_5\text{NH}_2$) is $93 \text{ g/mol}$.
Given mass of aniline = $9.3 \text{ g}$.
$ \text{Moles of Aniline} = \frac{9.3 \text{ g}}{93 \text{ g/mol}} = 0.1 \text{ mol} $
Step 2: Determine Theoretical Yield of Acetanilide
Approximate molar mass of acetanilide ($\text{C}_6\text{H}_5\text{NHCOCH}_3$) is $135 \text{ g/mol}$.
From the 1:1 stoichiometry, 0.1 mol of aniline theoretically yields 0.1 mol of acetanilide.
$ \text{Theoretical Yield} = \text{Moles} \times \text{Molar Mass} $
$ \text{Theoretical Yield} = 0.1 \text{ mol} \times 135 \text{ g/mol} = 13.5 \text{ g} $
Step 3: Calculate Percent Yield
The actual yield of acetanilide obtained is $11 \text{ g}$.
The formula for percent yield is:
$ \text{% Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% $
Substitute the values:
$ \text{% Yield} = \left( \frac{11 \text{ g}}{13.5 \text{ g}} \right) \times 100\% $
$ \text{% Yield} \approx 0.8148 \times 100\% \approx 81.48\% $
Rounding to one decimal place, the percent yield is $81.5\%$.
Similar Questions
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Important Questions from Physical Chemistry
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(Given: $\text{K}_{sp}(\text{XS}) = 1 \times 10^{-22}$ at $25^\circ\text{C}$, $\text{K}_{sp}(\text{YS}) = 4 \times 10^{-16}$ at $25^\circ\text{C}$, $[\text{H}_2\text{S}] = 0.1\text{M}$ in solution, $\text{K}_{a1} \times \text{K}_{a2}(\text{H}_2\text{S}) = 1.0 \times 10^{-21}$, $\log 2 = 0.30$, $\log 3 = 0.48$, $\log 5 = 0.70$) - Electricity is passed through an acidic solution of $\text{Cu}^{2+}$ till all the $\text{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ______ mL. (Nearest integer)
[Given:
$\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)} \quad \text{E}^\circ_{\text{red}} = +0.34 \text{ V}$
$\text{O}_2\text{(g)} + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O} \quad \text{E}^\circ_{\text{red}} = +1.23 \text{ V}$
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Molar mass of $\text{O}_2$ = $32 \text{ g mol}^{-1}$
Faraday Constant = $96500 \text{ C mol}^{-1}$
Molar volume at STP = 22.4 L] - Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and $15 \text{ kN m}^{-2}$ respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?
- At 298 K, the mole percentage of $\text{N}_2\text{(g)}$ in air is 80%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of $\text{N}_2\text{(g)}$ in water at 298 K?
($K_H \text{ for } \text{N}_2 \text{ is } 6.5 \times 10^7 \text{ mm Hg}$)
