Answer:
The percentage yield of the reaction is 37.24%.
Step-by-step explanation:
Experimental yield of nitrotoluene product = 305 g
Theoretical yield of nitrotoluene product = ?
Percentage yield:
![\% yield =\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100](https://img.qammunity.org/2020/formulas/chemistry/middle-school/fl1psq1d226l1xhc2bn2uucmoss57yyfyb.png)
Theoretical yield of nitrotoluene product:
![CH_3-C_6H_5+HNO_3\rightarrow CH_3-C_6H_4-NO_2+H_2O](https://img.qammunity.org/2020/formulas/chemistry/middle-school/h6d0v3x1x0vyc6z4ptjxjhln65qkibenzx.png)
Moles of toluene =
![(550 g)/(92 g/mol)=5.9783 mol](https://img.qammunity.org/2020/formulas/chemistry/middle-school/a92cizc09ae57e4zerj7uqcyrdqi83jhod.png)
According to reaction 1 mole of toluene gives 1 mole of nitrotoluene product.
Then 5.9783 moles of toluene will give:
of nitrotoluene.
Mass of 5.9783 moles of nitrotoluene:
5.9783 mol × 137 g/mol = 819.03 g
Percentage yield of the reaction:
![\% yield =\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100](https://img.qammunity.org/2020/formulas/chemistry/middle-school/fl1psq1d226l1xhc2bn2uucmoss57yyfyb.png)
![\% yield =(305 g)/(819.03 g)* 100=37.24\%](https://img.qammunity.org/2020/formulas/chemistry/middle-school/tydeyjlrz0ewkio29t1ykr2h39o03n9ns9.png)