Question

Given 10.5 g of ethane gas (C₂ H6) reacting with 26.0 L of oxygen gas, at STP,according to the following reaction.2 C₂H6+702 --> 4CO₂ + 6H₂OA) Calculate the mass of carbon dioxide gas produced. (3 marks)B) Determine the mass of excess reactant that remains. (2 marks)Show your work

Answer 1

Step-by-step explanation

Given:

Mass of ethane gas (C2H6) = 10.5g

Volume of Oxygen gas = 26.0 L

At STP : Pressure = 1 atm

: Temperature = 273 K

2C₂H6+702 --> 4CO₂ + 6H₂O

We know the molar mass of ethane = 30,07 g/mol

We know the molar mass of CO2 = 44.01 g/mol

Solution

A) Step 1: Find the moles of ethane

n = m/M where n is the moles, m is the mass and M is the molar mass

n = 10.5g/30,07 g/mol

n = 0.349 mol

Step 2: Use the stoichiometry to find the moles of carbon dioxide

The molar ratio between C2H6 and CO2 is 2:4

Therefore the moles of CO2 = 0.349 x (4/2)

= 0.698 mol

Step 3: From the moles of CO2, find the mass

m = n x M

m = 0.698 mol x 44.01 g/mol

m = 30.735 g

B) Step 1: Calculate the moles of oxygen, using the ideal gas law (R = 0.082 L.atm/K.mol)

PV = nRT

n = PV/RT

n = (1 atm x 26.0 L)/(0.082 L.atm/K.mol x 273K)

n = 1.161 mol

Step 2: Calculate how much moles will react with ethane to produce CO2

Moles of CO2 = 1.161 mol x (4/7)

Moles of CO2 = 0.663 mol

O2 produces more CO2, therefore O2 is the limiting reactant and C2H6 is the reactant in excess

mass of CO2 = n x M

= 0.663 mol x 44.01 g/mol

= 29.197 g

Step 3: Calculate the mass of C2H6 in excess

Use the O2 to find the moles of C2H6 that will react, then use those moles to find the mass.

Moles of C2H6 = 1.161 mol x (2/7)

= 0.332 mol

Mass of C2H6 = 0.332 mol x 30,07 g/mol

= 9.97 g

Step 4: Calculate the mass in excess

mass in excess = 10.5g - 9.97g = 0.53 g

Answers

A) Mass of CO2 = 29.197 g

B) Mass of O2 (excess reactant) in excess = 0.53g