Consider the reaction shown below. The percen | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The reaction is: $C_6H_5COCl + (C_6H_5)_2NH \rightarrow C_6H_5CON(C_6H_5)_2 + HCl$Molar mass of $C_6H_5COCl = 77 + 12 + 16 + 35.5 = 140.5 \, g/mol

Vedclass · Accepted Answer

$77$

Vedclass · Answer

(C) The reaction is: $C_6H_5COCl + (C_6H_5)_2NH ightarrow C_6H_5CON(C_6H_5)_2 + HCl$Molar mass of $C_6H_5COCl = 77 + 12 + 16 + 35.5 = 140.5 \, g/mol$.Molar mass of $C_6H_5CON(C_6H_5)_2 = 77 + 12 + 16 + 14 + 2(77) = 273 \, g/mol$.Moles of $C_6H_5COCl = \frac{0.140 \, g}{140.5 \, g/mol} \approx 0.001 \, mol$.Since the stoichiometry is $1:1$,theoretical moles of product $= 0.001 \, mol$.Theoretical mass of product $= 0.001 \, mol 	imes 273 \, g/mol = 0.273 \, g$.Actual mass of product formed $= 0.210 \, g$.Percentage yield $= \frac{	ext{Actual yield}}{	ext{Theoretical yield}} 	imes 100 = \frac{0.210}{0.273} 	imes 100 \approx 76.92 \%$.Rounding off to the nearest integer,we get $77 \%$.

Similar Questions

$28 \ g$ $KOH$ is required to completely neutralize $CO_2$ produced on heating $60 \ g$ of impure $CaCO_3$. The percentage purity of $CaCO_3$ is approximately (molar masses of $KOH$ and $CaCO_3$ are $56$ and $100 \ g \ mol^{-1}$, respectively).

To change the strength of $25 \, mL$ of $0.15 \, M$ $HCl$ to $0.1 \, M$,the quantity of water that is to be added is.......$mL$

What volume of a solution of hydrochloric acid containing $73 \ g$ of acid per litre would be sufficient for the exact neutralization of sodium hydroxide obtained by allowing $0.46 \ g$ metallic sodium to act upon water? $(Cl = 35.5, Na = 23.0, O = 16, H = 1)$

The volume of $0.6 \, M$ $NaOH$ required to neutralize $30 \, cm^3$ of $0.4 \, M$ $HCl$ is ........ $cm^3$.

How many moles of methane are required in the combustion reaction of methane to produce $22 \ g$ of $CO_{2(g)}$ (in $mol$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module