Question

Gaseous ethane

CH3CH3

reacts with gaseous oxygen gas

O2

to produce gaseous carbon dioxide

CO2

and gaseous water

H2O

. What is the theoretical yield of carbon dioxide formed from the reaction of

1.80g

of ethane and

12.5g

of oxygen gas?

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

Answer 1

Molar mass of C2H6,

MM = 2*MM(C) + 6*MM(H)

= 2*12.01 + 6*1.008

= 30.068 g/mol

mass(C2H6)= 1.80 g

number of mol of C2H6,

n = mass of C2H6/molar mass of C2H6

=(1.80 g)/(30.068 g/mol)

= 5.986*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 12.5 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(12.5 g)/(32 g/mol)

= 0.3906 mol

Balanced chemical equation is:

2 C2H6 + 7 O2 ---> 4 CO2 + 6 H2O

2 mol of C2H6 reacts with 7 mol of O2

for 0.059864 mol of C2H6, 0.209525 mol of O2 is required

But we have 0.390625 mol of O2

so, C2H6 is limiting reagent

we will use C2H6 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 = (4/2)* moles of C2H6

= (4/2)*0.059864

= 0.1197 mol

mass of CO2 = number of mol * molar mass

= 0.1197*44.01

= 5.27 g

Answer: 5.27 g