contestada

Consider the nitration by electrophilic aromatic substitution of salicylamide to iodosalicylamide. Reaction scheme illustrating the iodination of salicylamide by sodium iodide and sodium hypochlorite via an electrophilic aromatic substitution to form iodo-salicylamide. The density of salicylamide, d = 1.09 g/mL. A reaction was performed in which 3.65 mL of salicylamide was reacted with a mixture of concentrated nitric and sulfuric acids to make 5.33 g of iodosalicylamide. Calculate the theoretical yield and percent yield for this reaction.

Respuesta :

Answer: The percent yield of the reaction is 68.68%.

Explanation:

To calculate the mass of salicylamide, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of salicylamide = 1.06 g/mL

Volume of salicylamide = 3.65 mL

Putting values in above equation, we get:

[tex]1.06g/mL=\frac{\text{Mass of salicylamide}}{3.65mL}\\\\\text{Mass of salicylamide}=(1.06g/mL\times 3.65mL)=3.869g[/tex]

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 salicylamide = 3.869 g

Molar mass of salicylamide = 137.14 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of salicylamide}=\frac{3.869g}{137.14g/mol}=0.0295mol[/tex]

The chemical equation for the conversion of salicylamide to iodo-salicylamide follows:

[tex]\text{salicylamide }+NaI+NaOCl+EtOH\rightarrow \text{iodo-salicylamide }[/tex]

By Stoichiometry of the reaction:

1 mole of salicylamide produces 1 mole of iodo-salicylamide

So, 0.0295 moles of salicylamide will produce = [tex]\frac{1}{1}\times 0.0295=0.0295moles[/tex] of iodo-salicylamide

Now, calculating the mass of iodo-salicylamide from equation 1, we get:

Molar mass of iodo-salicylamide = 263 g/mol

Moles of iodo-salicylamide = 0.0295 moles

Putting values in equation 1, we get:

[tex]0.0295mol=\frac{\text{Mass of iodo-salicylamide}}{263g/mol}\\\\\text{Mass of iodo-salicylamide}=(0.0295mol\times 263g/mol)=7.76g[/tex]

To calculate the percentage yield of iodo-salicylamide, we use the equation:

[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield of iodo-salicylamide = 5.33 g

Theoretical yield of iodo-salicylamide = 7.76 g

Putting values in above equation, we get:

[tex]\%\text{ yield of iodo-salicylamide}=\frac{5.33g}{7.76g}\times 100\\\\\% \text{yield of iodo-salicylamide}=68.68\%[/tex]

Hence, the percent yield of the reaction is 68.68%.

Cricetus

The theoretical yield be "7.63 grams" and percent yield be "74.4%".

Percent and Theoretical yield:

According to the question,

Density, d = 1.09 g/mL

As we know the formula,

→ Density = [tex]\frac{Mass}{Volume}[/tex]

or,

      Mass = Density × Volume

Now,

Mass of Salicylamide reacted will be:

= 3.65 × 1.09

= 3.98 g

Moles of Salicylamide will be:

= [tex]\frac{3.98}{137.14}[/tex]

= 0.029 mol

Now,

0.029 mol of salicylamide will form product = 0.029 mol

hence,

The theoretical yield will be:

= 0.029 × 263.03

= 7.63 grams

and,

The percent yield be:

= [tex]\frac{5.68}{7.63}[/tex]

= 74.4 %

Thus the above responses are correct.

Find out more information about theoretical yield here:

https://brainly.com/question/25996347

Otras preguntas