Question

For the following reaction, 3.80 grams of hydrogen gas are allowed to react with 10.6 grams of ethylene (C2H4). hydrogen (g) + ethylene (C2H4) (g) ethane (C2H6) (g)

What is the maximum amount of ethane (C₂H₆) that can be formed?

What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete

Answer 1

1)

Molar mass of C2H4,

MM = 2*MM(C) + 4*MM(H)

= 2*12.01 + 4*1.008

= 28.052 g/mol

mass(C2H4)= 10.6 g

number of mol of C2H4,

n = mass of C2H4/molar mass of C2H4

=(10.6 g)/(28.052 g/mol)

= 0.3779 mol

Molar mass of H2 = 2.016 g/mol

mass(H2)= 3.8 g

number of mol of H2,

n = mass of H2/molar mass of H2

=(3.8 g)/(2.016 g/mol)

= 1.885 mol

Balanced chemical equation is:

C2H4 + H2 ---> C2H6 +

1 mol of C2H4 reacts with 1 mol of H2

for 0.3779 mol of C2H4, 0.3779 mol of H2 is required

But we have 1.8849 mol of H2

so, C2H4 is limiting reagent

we will use C2H4 in further calculation

Molar mass of C2H6,

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

= 2*12.01 + 6*1.008

= 30.068 g/mol

According to balanced equation

mol of C2H6 formed = (1/1)* moles of C2H4

= (1/1)*0.3779

= 0.3779 mol

mass of C2H6 = number of mol * molar mass

= 0.3779*30.07

= 11.4 g

Answer: 11.4 g

2)

C2H4 is limiting reagent

3)

According to balanced equation

mol of H2 reacted = (1/1)* moles of C2H4

= (1/1)*0.3779

= 0.3779 mol

mol of H2 remaining = mol initially present - mol reacted

mol of H2 remaining = 1.8849 - 0.3779

mol of H2 remaining = 1.5071 mol

Molar mass of H2 = 2.016 g/mol

mass of H2,

m = number of mol * molar mass

= 1.507 mol * 2.016 g/mol

= 3.04 g

Answer: 3.04 g of H2 remains