Respuesta :
Answer: The theoretical yield of 4-nitrochalcone is, [tex]2.13\times 10^2[/tex]
Explanation : Given,
Volume of acetophenone = 135 microliters = 135 × 10⁻⁶ L = 0.135 mL
conversion used : (1 microliter = 10⁻⁶ L) and (1 L = 1000 mL)
Density of acetophenone = 1.03 g/mL
Mass of acetophenone = Density × Volume = 1.03 g/mL × 0.135 mL = 0.139 g
Mass of 4-nitrobenzaldehyde = 127 mg = 0.127 g
Conversion used : (1 mg = 0.001 g)
First we have to calculate the moles of acetophenone and 4-nitrobenzaldehyde
[tex]\text{Moles of acetophenone}=\frac{\text{Given mass acetophenone}}{\text{Molar mass acetophenone}}[/tex]
[tex]\text{Moles of acetophenone}=\frac{0.139g}{120.15g/mol}=0.00116mol[/tex]
and,
[tex]\text{Moles of 4-nitrobenzaldehyde}=\frac{\text{Given mass 4-nitrobenzaldehyde}}{\text{Molar mass 4-nitrobenzaldehyde}}[/tex]
[tex]\text{Moles of 4-nitrobenzaldehyde}=\frac{0.127g}{151.12g/mol}=0.000840mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]C_8H_8O+C_7H_5NO_3\rightarrow C_{15}H_{11}NO_3[/tex]
From the balanced reaction we conclude that
As, 1 mole of 4-nitrobenzaldehyde react with 1 mole of acetophenone
So, 0.000840 mole of 4-nitrobenzaldehyde react with 0.000840 mole of acetophenone
From this we conclude that, acetophenone is an excess reagent because the given moles are greater than the required moles and 4-nitrobenzaldehyde is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of 4-nitrochalcone
From the reaction, we conclude that
As, 1 mole of 4-nitrobenzaldehyde react to give 1 mole of 4-nitrochalcone
So, 0.000840 mole of 4-nitrobenzaldehyde react to give 0.000840 mole of 4-nitrochalcone
Now we have to calculate the mass of 4-nitrochalcone
[tex]\text{ Mass of 4-nitrochalcone}=\text{ Moles of 4-nitrochalcone}\times \text{ Molar mass of 4-nitrochalcone}[/tex]
Molar mass of 4-nitrochalcone = 253.25 g/mole
[tex]\text{ Mass of 4-nitrochalcone}=(0.000840moles)\times (253.25g/mole)=0.21273g=212.73mg=2.13\times 10^2mg[/tex]
(1 g = 1000 g)
Therefore, the theoretical yield of 4-nitrochalcone is, [tex]2.13\times 10^2mg[/tex]
