The ethane will react with oxygen to produce carbon dioxide and water:
2 C₂H₆ + 7 O₂ ----> 4 CO₂ + 6 H₂O
We have to find the volume of water produced (we will consider that it is vaporized) from 45 g of oxygen with excess ethane. The first step is to convert those grams into moles of oxygen using the molar mass of it.
atomic mass of O = 16.00 amu
molar mass of O₂ = 2 * 16.00 g/mol
molar mass of O₂ = 32 g/mol
moles of O₂ = mass of O₂/molar mass of O₂
moles of O₂ = 45 g/(32 g/mol)
moles of O₂ = 1.41 moles
2 C₂H₆ + 7 O₂ ----> 4 CO₂ + 6 H₂O
According to the balanced equation, when 7 moles of O₂ (reacting with excess ethane) produces 6 moles of H₂O. So we can use that relationship to find the number of moles of H₂O produced by 1.41 moles of O₂.
moles of H₂O = 1.41 moles of O₂ * 6 moles of H₂O/(7 moles of O₂)
moles of H₂O = 1.21 moles
Now we can use the ideal gas law to find the volume of water.
P * V = n * R * T
R = 0.082 atm*L/(mol*K)
P = 742 torr
1 atm = 760 torr
P = 742 torr * 1 atm/(760 torr)
P = 0.976 atm
V = ?
n = 1.21 moles
T = 35 °C = (35 + 273.15) K
T = 308.15 K
We will find the answer to our problem by replacing those values in the ideal gas law and solving it for the volume:
P * V = n * R * T
V = n * R * T /P
V = 1.21 mol * 0.082 atm*L/(mol*K) * 308.15 K/(0.976 atm)
V = 31.3 L
Answer: 31.3 L of water is produced.