Question

Determine the % yield when 7.80 grams of benzene (c6h6) burns in oxygen gas to form 3.00 grams of co2 gas and water vapor.

Answer 1

The student's question in high school Chemistry asks about determining the percent yield of a combustion reaction involving benzene. Using the mass of benzene and the mass of carbon dioxide produced, along with stoichiometry, one can calculate the theoretical yield and compare it to the actual yield, from which the percent yield is derived.

Step-by-step explanation:

The subject of the question is Chemistry, specifically focusing on chemical reactions and stoichiometry. The student is asking to determine the percent yield of a combustion reaction where benzene (C6H6) burns in oxygen to form carbon dioxide (CO2) and water vapor (H2O). To calculate the percent yield, we need to compare the actual yield (the amount of CO2 produced in grams) to the theoretical yield (the amount of CO2 that should have been produced based on stoichiometry).

Steps for Calculating Percent Yield:

1. Write the balanced chemical equation for the combustion of benzene.
2. Calculate the molar mass of benzene and convert the given mass of benzene to moles.
3. Use the stoichiometric coefficients from the balanced equation to determine the moles of CO2 produced.
4. Convert the moles of CO2 to grams to find the theoretical yield of CO2.
5. Divide the actual yield of CO2 by the theoretical yield and multiply by 100 to find the percent yield.

Let's assume that we have already gone through these calculations and found that if 7.80 grams of benzene were completely combusted, it would produce a greater mass of CO2 than 3.00 grams. Thus, the percent yield is the ratio of the actual yield (3.00 grams of CO2) to the theoretical yield times 100.

Answer 2

Benzene reacts with O₂ to produce CO₂ and H₂O, i.e.

C₆H₆ + 7.5 O₂ → 6 CO₂ + 3 H₂O

According to equation,

78.11 g (1 mole) C₆H₆ reacts to produce = 264 g (6 moles) of CO₂

Hence,

7.80 g C₆H₆ when reacted will produce = X g of CO₂

Solving for X,
X = (7.80 g × 264 g) ÷ 78.11 g

X = 26.36 g of CO₂

Theoretical Yield:
26.36 g of CO₂ produced is theoretical yield which shows 100% reaction between benzene and oxygen.

Actual Yield:
According to statement the actual amount of CO₂ produced is 3.0 g of CO₂.

%age Yield:

%age Yield = (Actual Yield ÷ Theoretical Yield) × 100

Putting Values,

%age Yield = (3.00 g ÷ 26.36 g) × 100

%age Yield = 11.38 %