Question

Gaseous butane CH3CH22CH3

reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 7.47g of carbon dioxide is produced from the reaction of 4.1g of butane and 9.8g of oxygen gas, calculate the percent yield of carbon dioxide.

Be sure your answer has the correct number of significant digits in it.

Answer 1

Molar mass of C4H10,

MM = 4*MM(C) + 10*MM(H)

= 4*12.01 + 10*1.008

= 58.12 g/mol

mass(C4H10)= 4.1 g

number of mol of C4H10,

n = mass of C4H10/molar mass of C4H10

=(4.1 g)/(58.12 g/mol)

= 7.054*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 9.8 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(9.8 g)/(32 g/mol)

= 0.3063 mol

Balanced chemical equation is:

2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O

2 mol of C4H10 reacts with 13 mol of O2

for 0.0705 mol of C4H10, 0.4585 mol of O2 is required

But we have 0.3063 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

According to balanced equation

mol of CO2 formed = (8/13)* moles of O2

= (8/13)*0.3063

= 0.1885 mol

mass of CO2 = number of mol * molar mass

= 0.1885*44.01

= 8.294 g

% yield = actual mass*100/theoretical mass

= 7.47*100/8.294

= 90.1 %

Answer: 90.1 %