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damaskus [11]
3 years ago
15

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.
Chemistry
1 answer:
kvv77 [185]3 years ago
4 0

<u>Answer:</u> The percent yield of the reaction is 68.68%.

<u>Explanation:</u>

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

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of salicylamide = 1.06 g/mL

Volume of salicylamide = 3.65 mL

Putting values in above equation, we get:

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

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of salicylamide = 3.869 g

Molar mass of salicylamide = 137.14 g/mol

Putting values in equation 1, we get:

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

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

\text{salicylamide }+NaI+NaOCl+EtOH\rightarrow \text{iodo-salicylamide }

By Stoichiometry of the reaction:

1 mole of salicylamide produces 1 mole of iodo-salicylamide

So, 0.0295 moles of salicylamide will produce = \frac{1}{1}\times 0.0295=0.0295moles 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:

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

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

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iodo-salicylamide = 5.33 g

Theoretical yield of iodo-salicylamide = 7.76 g

Putting values in above equation, we get:

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

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

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Explanation:

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Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}

E^o_{cell}=-0.036V-(-2.37 V)=2.334 V

The overall reaction will be:

2 × [1] + 3 × [2] :

2Fe^{3+} (aq) + 3Mg(s)+6e^-\rightarrow 2Fe (s)+3Mg^{2+}(aq)+6e^-

Electrons on both sides will get cancelled :

2Fe^{3+} (aq) + 3Mg(s)\rightarrow 2Fe (s)+3Mg^{2+}(aq)

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Which condition must be met in order for an equation to be balanced?
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              # of Moles  =  Number of Atoms / 6.022 × 10²³
Or,
             Number of Atoms  =  Moles × 6.022 × 10²³     ------- (1)    

Calculating Moles,
As,
                              Moles  =  Mass / M.mass
So,
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Putting value of mole in eq.1,

             Number of Atoms  =  0.0215 × 6.022 × 10²³

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Result:
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