Respuesta :
Answer: The percent yield of the reaction is 68.16 %.
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 camphor:
Given mass of camphor = 1.500 g
Molar mass of camphor = 152.23 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of camphor}=\frac{1.500g}{152.23g/mol}=9.85\times 10^{-3}mol[/tex]
The chemical equation for the reaction of camphor and sodium borohydride follows:
[tex]\text{Camphor}+NaBH_4\rightarrow \text{Isoborneol}[/tex]
As, sodium borohydride is present in excess. It is an excess reagent. So, camphor is the limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
1 mole of camphor produces 1 mole of isoborneol
So, [tex]9.85\times 10^{-3}mol[/tex] of camphor will produce = [tex]\frac{1}{1}\times 9.85\times 10^{-3}mol=9.85\times 10^{-3}mol[/tex] of isoborneol
- Now, calculating the mass of isoborneol from equation 1, we get:
Molar mass of isoborneol = 154.25 g/mol
Moles of isoborneol = [tex]9.85\times 10^{-3}[/tex] moles
Putting values in equation 1, we get:
[tex]9.85\times 10^{-3}mol=\frac{\text{Mass of isoborneol}}{154.25g/mol}\\\\\text{Mass of isoborneol}=(9.85\times 10^{-3}mol\times 154.25g/mol)=1.52g[/tex]
- To calculate the percentage yield of isoborneol, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of isoborneol = 1.036 g
Theoretical yield of isoborneol = 1.52 g
Putting values in above equation, we get:
[tex]\%\text{ yield of isoborneol}=\frac{1.036g}{1.52g}\times 100\\\\\% \text{yield of isoborneol}=68.16\%[/tex]
Hence, the percent yield of the reaction is 68.16 %.
