Question

How much CO2 can be produced in the reaction of 0.29 mole of C2H3Br3 and 1.53 moles O2?

4 C2H3Br3 + 11 O2 = 8 CO2 + 6 H2O + 6 Br2



a. Identify the limiting reactant.

b. Identify the excess reactant.

c. What amount of the excess reactant is left over after the reaction?

d. What is the theoretical yield?

e. Compute for the percent yield​

Answer 1

the answer to this question is 6Br2

Answer 2

Answer:

The limiting reactant is $C_{2}H_{3}Br_{3}$ and the excess reactant is $O_{2}$.

Moles of $C_{2}H_{3}Br_{3}$ = 0.29 moles

Moles of $O_{2}$ = 1.53 moles

Divide the moles of each reactant by the corresponding stoichiometric coefficient:

The stoichiometric coefficient of $C_{2}H_{3}Br_{3}$ is 4 and that of $O_{2}$ is 11. Therefore,

0.29/4 = 0.0725 moles of $C_{2}H_{3}Br_{3}$  

1.53/11 = 0.139 moles of $O_{2}$  

As the moles of $C_{2}H_{3}Br_{3}$ is less than the moles of $O_{2}$, $C_{2}H_{3}Br_{3}$ is the limiting reactant and $O_{2}$ is the excess reactant.

The moles of $O_{2}$ that will react with 0.29 moles of $C_{2}H_{3}Br_{3}$ are (0.29/4)*11 = 0.798 moles.

Amount of excess reactant left  = 1.53 - 0.798 = 0.732 moles

Therefore, the amount of excess reactant left over after the reaction is 0.732 moles.

The ratio of molecules in product and reactant is 8/4 = 2. That is, for every molecule of $C_{2}H_{3}Br_{3}$, two molecules of $CO_{2}$ can be produced.

Theoretical yield = moles of  $C_{2}H_{3}Br_{3}$ × ratio of molecules × molar mass

Theoretical yield = 0.29 × 2 × 44

Theoretical yield = 25.52g

Therefore, the theoretical yield obtained is 25.52 grams of $CO_{2}$.

Percentage yield = (actual yield / theoretical yield) × 100

Percentage yield = (2/25.52) × 100

Percentage yield = 7.84%

Therefore, the percentage yield obtained is 7.84%.

#SPJ2