Verified Solution • 2025–26 Session moderate QJEEM25Cha192: JEE Main 2025 Chemistry - Class 11 JEEM25Cha192 Solution: Xg of benzoic acid on reaction with aq released that occupied… Xg of benzoic acid on reaction with aq NaHCO3\mathrm{NaHCO}_3NaHCO3 released CO2\mathrm{CO}_2CO2 that occupied 11.2 L volume at STP. X is _________ g. Step-by-Step Verified Solution amp;C6H5COOH+NaHCO3→C6H5COONa+amp;+H2O+CO2\begin{aligned} & \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}+\mathrm{NaHCO}_3 \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{COO} \mathrm{Na}^{+} \\ & +\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\end{aligned}amp;C6H5COOH+NaHCO3→C6H5COONa+amp;+H2O+CO2 amp;x gm11.2 Lamp; mole of C6H5COOH= mole of CO2=11.222.4=0.5amp; mass of C6H5COOH=x=0.5×122=61gm\begin{aligned}& \mathrm{x} \mathrm{~gm} \quad \text{11.2 L}\\& \text { mole of } \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}=\text { mole of } \mathrm{CO}_2=\frac{11.2}{22.4}=0.5 \\& \text { mass of } \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}=\mathrm{x}=0.5 \times 122=61 \mathrm{gm}\end{aligned}amp;x gm11.2 Lamp; mole of C6H5COOH= mole of CO2=22.411.2=0.5amp; mass of C6H5COOH=x=0.5×122=61gm Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring