Question

a. If 42.5 g of CH3OH reacts with 22.8 L of O2 at 27°C and a pressure of 2.00 atm, calculate the number of grams of water vapor and the volume of water vapor that would be formed at the same temperature and pressure.​

Answer 1

The mass of water vapor is 44.46 grams

The volume of water is 30.37 L

Step-by-step 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