Answer:
The mass of water vapor is 44.46 grams
The volume of water is 30.37 L
Explanation:
Step 1: Data given
Mass of CH3OH =42.5 grams
Molar mass CH3OH = 32.04 g/mol
Volume of O2 = 22.8 L
Pressure = 2.00 atm
Step 2: The balanced equation
2CH3OH + 3O2 → 2CO2 + 4H2O
Step 3: Calculate moles CH3OH
Moles CH3OH = mass CH3OH / molar mass CH3OH
Moles CH3OH = 42.5 grams / 32.04 g/mol
Moles CH3OH = 1.326 moles
Step 4: Calculate moles O2
p*V = n*R*T
⇒with p = the pressure = 2.00 atm
⇒with V = the volume of O2 = 22.8 L
⇒with n = the moles of O2 = TO BE DETERMINED
⇒with R = the gas constant = 0.08206 L*Atm/mol*K
⇒with T = the temperature = 27 °C = 300 K
n = (p*V) / (R*T)
n = (2.00 * 22.8) / (0.08206*300)
n = 1.85 moles
Step 5: Calculate the limiting reactant
For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2 and 4 H2O
O2 is the limiting reactant. It will completely be consumed ( 1.85 moles). CH3OH is in excess. There will react 2/3*1.85 = 1.233 moles. There will remain 1.326 - 1.233 = 0.093 moles
Step 6: Calculate moles products
For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2 and 4 H2O
For 1.85 moles O2 we'll have 1.233 moles CO2 and 2.467 moles H2O
Step 7: Calculate mass H2O
Mass H2O = moles H2O * molar mass H2O
Mass H2O = 2.467 moles * 18.02 g/mol
Mass H2O = 44.46 grams
Step 8: Calculate volume H2O
p*V = n*R*T
⇒with p = the pressure = 2.00 atm
⇒with V = the volume of H2O = TO BE DETERMINED
⇒with n = the moles of H2O = 2.467 moles
⇒with R = the gas constant = 0.08206 L*Atm/mol*K
⇒with T = the temperature = 27 °C = 300 K
V = (n*R*T)/p
V = (2.467 * 0.08206 * 300) / 2.00
V = 30.37 L
The mass of water vapor is 44.46 grams
The volume of water is 30.37 L
