Question

A rigid 7.80 L sealed vessel containing 2.700 mol of O2(g), 0.600 mol of C2H4(g), and 2.400 mol of Ne(g) has an internal temperature of 85.0∘C
a. Calculate the total pressure in the cylinder.
b. Find the mole fraction of O2 in the vessel.
c. A lab technician ignites the mixture in the vessel and the following reaction occurs:

C2H4(g)+3O2(g)→2CO2(g)+2H2O(g)

Find the mole fraction of each gas in the vessel after the reaction.

Answer 1

Answer:

a. 21.46atm

b. 0.4737

c.

Mole fraction O2 = 0.1579

Mole fraction CO2 = 0.2105

Mole fraction H2O 0.2105

Mole fraction Ne = 0.4211

Explanation:

a. To solve the total pressure of the cylinder we need to use:

PV = nRT

Where P is pressure: Our incognite

V is volume: 7.80L

n are total moles: 2.700mol + 0.600mol + 2.400mol = 5.700moles

R is gas constant: 0.082atmL/molK

And T is absolute temperature in K: 85°C + 273.15 = 358.15K

P = nRT/V

P = 5.700mol*0.082atmL/molK*358.15K/7.80L

P = 21.46atm

b. Mole fraction of O2 is the ratio of moles of O2 (2.700mol) over total moles of the mixture (5.700mol).

Mole fraction = 2.700mol / 5.700mol = 0.4737

c. In the reaction:

C2H4(g)+3O2(g)→2CO2(g)+2H2O(g)

For a complete reaction of the 0.600mol of C2H4 there are required:

0.600mol C2H4 * (3 mol O2 / 1mol C2H4) = 1.800 moles of O2

As there are 2.700moles of O2, will remain:

2.700 - 1.800 = 0.900moles of O2

All C2H4 is consumed because is limitng reactant and there are produced:

0.600mol C2H4 * (2 mol CO2 / 1mol C2H4) = 1.200 moles of CO2

0.600mol C2H4 * (2 mol H2O / 1mol C2H4) = 1.200 moles of H2O

And 2.400 mol of Ne that doesn't react.

Total moles are:

0.900mol + 1.200mol + 1.200mol + 2.400mol = 5.700mol

Mole fraction O2 = 0.900mol / 5.700mol = 0.1579

Mole fraction CO2 = 1.200mol / 5.700mol = 0.2105

Mole fraction H2O = 1.200mol / 5.700mol = 0.2105

Mole fraction Ne = 2.400mol / 5.700mol = 0.4211