Respuesta :
Answer: The percent yield of the reaction is 32.04 %.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of benzoic acid = 3.6 g
Molar mass of benzoic acid = 112.12 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of benzoic acid}=\frac{3.6g}{112.12g/mol}=0.0321mol[/tex]
The chemical equation for the formation of methyl benzoate from benzoic acid follows:
[tex]\text{Benzoic acid + Methanol}\rightarrow \text{Methyl benozate}[/tex]
By Stoichiometry of the reaction:
1 mole of benzoic acid produces 1 mole of methyl benzoate
So, 0.0321 moles of benzoic acid will produce = [tex]\frac{1}{1}\times 0.0321=0.0321mol[/tex] of methyl benzoate
Now, calculating the mass of methyl benzoate from equation 1, we get:
Molar mass of methyl benzoate = 136.15 g/mol
Moles of methyl benzoate = 0.0321 moles
Putting values in equation 1, we get:
[tex]0.0321mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0321mol\times 136.15g/mol)=4.37g[/tex]
To calculate the percentage yield of methyl benzoate, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of methyl benzoate = 1.4 g
Theoretical yield of methyl benzoate = 4.37 g
Putting values in above equation, we get:
[tex]\%\text{ yield of methyl benzoate}=\frac{1.4g}{4.37g}\times 100\\\\\% \text{yield of methyl benzoate}=32.04\%[/tex]
Hence, the percent yield of the reaction is 32.04 %.
