Question

Gaseous ethane CH3CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose 14. g of ethane is mixed with 37.6 g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction.

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)= 14.0 g

number of mol of C2H6,

n = mass of C2H6/molar mass of C2H6

=(14.0 g)/(30.068 g/mol)

= 0.4656 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 37.6 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(37.6 g)/(32 g/mol)

= 1.175 mol

Balanced chemical equation is:

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

2 mol of C2H6 reacts with 7 mol of O2

for 0.465611 mol of C2H6, 1.629639 mol of O2 is required

But we have 1.175 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

According to balanced equation

mol of C2H6 recated = (2/7)* moles of O2

= (2/7)*1.175

= 0.335714 mol

mol of C2H6 remaining = mol initially present - mol reacted

mol of C2H6 remaining = 0.465611 - 0.335714

mol of C2H6 remaining = 0.129897 mol

mass of C2H6,

m = number of mol * molar mass

= 0.1299 mol * 30.068 g/mol

= 3.91 g

Answer: 3.91 g