Question

Answer the following questions using the balanced chemical equation below:

Bi₂O₃ + 3 C ? 2 Bi + 3 CO

a. What is the limiting reactant when 232.47 g Bi₂O₃ reacts with 27.03 g C?

b. What is the mass of bismuth produced from this reaction (i.e. the theoretical yield)?


c. If the actual yield is 123.20 g Bi, what is the percent yield?


d. How much excess reagent is left?

Answer 1

Balanced equation:
Bi₂O₃ + 3 C ====> 2 Bi + 3 CO
Reaction type: single replacement

a. What is the limiting reactant when 232.47 g Bi2O3 reacts with 27.03 g C?

232.47 g Bi2O3 = 232.47/ 465.95 =  0.4989 Moles

27.03 g C = 27.03 / 12 =  2.250 Moles

limiting reactant is Bi2O3

b. What is the mass of bismuth produced from this reaction (i.e. the theoretical yield)?

Moles of Bi produced =  0.9978 Moles

Mass of Bi produced =  0.9978 x 208.98 =  208.52 gm

c. If the actual yield is 123.20 g Bi, what is the percent yield?

Thoretical yield = 208.52 gm

Actual yield = 123.20 gm

percent yield =  Actual yield x 100 / Thoretical yield  

percent yield = 123.20 x 100 / 208.52 = 59.08 %

d. How much excess reagent is left?

Moles of carbon used =  1.496 Moles

Mass of carbon used = 1.496 x 12 = 17.97 gm

Excess carbon left = 9.0535