5. consider the following reaction: (ℓ) + 4.5 o₂ (g) → + 2 co₂ (g) + 2 h₂o (g)\na) calculate the overall…

5. consider the following reaction: (ℓ) + 4.5 o₂ (g) → + 2 co₂ (g) + 2 h₂o (g)\na) calculate the overall atom efficiency (oae)\nb) calculate the e - factor\nc) in the lab you make 7.5 g, but the theoretical yield was 8 g. what is the % yield for this reaction?\nd) comment on the green nature of this reaction for the following reaction:

5. consider the following reaction: (ℓ) + 4.5 o₂ (g) → + 2 co₂ (g) + 2 h₂o (g)\na) calculate the overall atom efficiency (oae)\nb) calculate the e - factor\nc) in the lab you make 7.5 g, but the theoretical yield was 8 g. what is the % yield for this reaction?\nd) comment on the green nature of this reaction for the following reaction:

Answer

Explanation:

Step1: Calculate molar - masses

Let's assume the reactant (benzene) has a molar - mass (M_{benzene}=78\ g/mol), the product (maleic anhydride) has a molar - mass (M_{maleic\ anhydride}=98\ g/mol), (M_{CO_2}=44\ g/mol), and (M_{H_2O}=18\ g/mol).

Step2: Calculate the overall atom efficiency (OAE)

The formula for OAE is (OAE=\frac{\text{Mass of atoms in desired product}}{\text{Mass of atoms in all reactants}}\times100%). The mass of atoms in the desired product (maleic anhydride) is (m_{product}=98\ g/mol). The mass of atoms in all reactants: Benzene ((C_6H_6)) has a molar - mass of (78\ g/mol) and oxygen ((4.5\ mol) of (O_2)) has a mass of (4.5\times32 = 144\ g/mol). The total mass of reactants (m_{reactants}=78 + 144=222\ g/mol). (OAE=\frac{98}{222}\times100%\approx44.1%)

Step3: Calculate the E - factor

The E - factor is defined as (\text{E - factor}=\frac{\text{Mass of waste}}{\text{Mass of product}}). The mass of waste: (2\ mol) of (CO_2) has a mass of (2\times44 = 88\ g/mol) and (2\ mol) of (H_2O) has a mass of (2\times18 = 36\ g/mol). The total mass of waste (m_{waste}=88 + 36=124\ g/mol). The mass of the product (maleic anhydride) is (m_{product}=98\ g/mol). (\text{E - factor}=\frac{124}{98}\approx1.27)

Step4: Calculate the percentage yield

The formula for percentage yield is (%\text{Yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100%). Given actual yield (=7.5\ g) and theoretical yield ( = 8\ g). (%\text{Yield}=\frac{7.5}{8}\times100% = 93.75%)

Step5: Comment on the green nature

The OAE is relatively low ((44.1%)), indicating that a significant portion of the atoms in the reactants do not end up in the desired product. The E - factor of (1.27) means that for every gram of product, there is about (1.27) grams of waste. The reaction is not very green as it has low atom - efficiency and a non - negligible amount of waste.

Answer:

a) (44.1%) b) (1.27) c) (93.75%) d) The reaction is not very green due to low atom - efficiency ((44.1%)) and a non - negligible E - factor ((1.27)).