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