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PLEASE HELP!!!!!!
A student starts with 0.303 g of pure 4-aminobenzoic acid, (molar mass 137.14 g/mol) and ends up with 0.318 g of benzocaine (molar mass 165.19 g/mol).


Given that ethanol is in excess, what are the theoretical and percent yields for this reaction? Assume all starting material is converted to product when calculating the theoretical yield

PLEASE HELPA Student Starts With 0303 G Of Pure 4aminobenzoic Acid Molar Mass 13714 Gmol And Ends Up With 0318 G Of Benzocaine Molar Mass 16519 Gmol Given That class=

Sagot :

sebassandin

Answer:

[tex]m_{B}^{theoretical}=0.365gB[/tex]

[tex]Y=87.1\%[/tex]

Explanation:

Hello there!

In this case, since the reaction (A->B) have an initial amount of pure 4-aminobenzoic acid, the first step to compute the theoretical yield is to solve the following stoichiometric setup:

[tex]m_{B}^{theoretical}=0.303gA*\frac{1molA}{137.14gA}*\frac{1molB}{1molA}*\frac{165.19 gB}{1molB}\\\\ m_{B}^{theoretical}=0.365gB[/tex]

Whereas A stands for 4-aminobenzoic acid and B for the benzocaine. Moreover, we compute the percent yield by dividing the actual yield (0.318 g) by the theoretical one (0.365 g):

[tex]Y=\frac{0.318g}{0.365g} *100\%\\\\Y=87.1\%[/tex]

Best regards!

Eduard22sly

The percentage yield obtained when 0.303 g of 4-aminobenzoic acid react with excess ethanol is 87.1%

  • We'll begin by calculating the theoretical yield of benzocaine. This can be obtained as follow:

From the balanced equation given in the question above,

137.14 g of 4-aminobenzoic acid reacted to produce 165.19 g of benzocaine.

Therefore,

0.303 g of 4-aminobenzoic acid will react to produce =[tex]\frac{0.303 * 165.19}{137.14}[/tex] = 0.365 g of benzocaine.

Thus, the theoretical yield of benzocaine is 0.365 g

  • Finally, we shall determine the percentage yield. This can be obtained as follow:

Actual yield of benzocaine = 0.318 g

Theoretical yield of benzocaine = 0.365 g

Percentage yield =?

[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{0.318}{0.365} * 100\\\\[/tex]

= 87.1%

Therefore, the percentage yield for the reaction is 87.1%

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