Question

28.06 g of ethene (C2H4) gas is mixed with 158.72 g of oxygen (O2) gas. The mixture is ignited. After a bright flash and a loud bang, some water droplets form on the inside of the reaction vessel.

PART A

How Much Oxygen is left over?

PART B

Calculate the theoretical Yield of Water.

PART C

Calculate the theoretical yield of Carbon Dioxide.

Answer 1

A)

Molar mass of C2H4,

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

= 2*12.01 + 4*1.008

= 28.052 g/mol

mass(C2H4)= 28.06 g

use:

number of mol of C2H4,

n = mass of C2H4/molar mass of C2H4

=(28.06 g)/(28.05 g/mol)

= 1 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 158.72 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(1.587*10^2 g)/(32 g/mol)

= 4.96 mol

Balanced chemical equation is:

C2H4 + 3 O2 ---> 2 CO2 + 2 H2O

1 mol of C2H4 reacts with 3 mol of O2

for 1 mol of C2H4, 3.001 mol of O2 is required

But we have 4.96 mol of O2

so, C2H4 is limiting reagent

we will use C2H4 in further calculation

According to balanced equation

mol of O2 reacted = (3/1)* moles of C2H4

= (3/1)*1

= 3.001 mol

mol of O2 remaining = mol initially present - mol reacted

mol of O2 remaining = 4.96 - 3.001

mol of O2 remaining = 1.959 mol

Molar mass of O2 = 32 g/mol

use:

mass of O2 remaining,

m = number of mol * molar mass

= 1.959 mol * 32 g/mol

= 62.69 g

Answer: 62.7 g

B)

)

Molar mass of H2O,

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

= 2*1.008 + 1*16.0

= 18.016 g/mol

According to balanced equation

mol of H2O formed = (2/1)* moles of C2H4

= (2/1)*1

= 2.001 mol

use:

mass of H2O = number of mol * molar mass

= 2.001*18.02

= 36.0 g

Answer: 36.0 g

C)

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 = (2/1)* moles of C2H4

= (2/1)*1

= 2.001 mol

use:

mass of CO2 = number of mol * molar mass

= 2.001*44.01

= 88.05 g

Answer: 88.1 g