Respuesta :
Answer: The percent yield of the reaction is 32.34 %
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)
- For benzoic acid:
Given mass of benzoic acid = 3.6 g
Molar mass of benzoic acid = 122.12 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol[/tex]
The chemical equation for the reaction of benzoic acid and methanol is:
[tex]\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}[/tex]
By Stoichiometry of the reaction
1 mole of benzoic acid produces 1 mole of methyl benzoate
So, 0.0295 moles of benzoic acid will produce = [tex]\frac{1}{1}\times 0.0295=0.108[/tex] moles 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 = 0.0295 moles
Putting values in equation 1, we get:
[tex]0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g[/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 = 1.3 g
Theoretical yield of methyl benzoate = 4.02 g
Putting values in above equation, we get:
[tex]\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%[/tex]
Hence, the percent yield of the reaction is 32.34 %
