Question

2nd attempt The reaction of 9.85 g of carbon with excess O2 yields 10.2 g of CO2. What is the percent yield of this reaction? 1st attempt

Answer 1

Balanced chemical equation is

C + O2 $1585522755019_blob.png$ CO2

First calculate theoratical yield

molar mass of C = 12 g/mol

gm of C = 9.85 gm

no.of mole = gm of compound / molar mass

moles of C = 9.85 / 12 = 0.82083 mole

According to balanced reaction 1 mole C produce mole CO2 molar ratio between C to CO2 is 1:1 therefore 0.82083 mole of C produce 0.82083 mole of CO2

mole of CO2 = 0.82083 moles

molar mass of CO2 = 44.01 g/mol

gm of compound = no. of mole X molar mass

gm of  CO2​​​ = 0.82083 X 44.01 = ​​​​36.125 gm

theoratical yield = 36.125 gm

percent yield = actual yield X 100 / theoratical yield

Actual yield of  CO2​​​ = 10.2 g

substitute the value

percent yield of  CO2​​​ = 10.2 X 100 / 36.125 = 28.24 %

percent yield for the reaction = 28.24 %