Answer:
A: There is 43.12 Liter of H2 needed
B: There is 21.56 liter of CO needed
Explanation:
Step 1: Data given
CO(g)+2H2(g)?CH3OH(g)
For 1 mol of CO consumed, we need 2 moles of H2 to produce 1 mol of CH3OH
Molar mass of CO = 28.01 g/mol
Molar mass of H2 = 2.02 g/mol
Molar mass of CH3OH = 32.04 g/mol
Step 2: What volume of H2 gas (in L), measured at 754mmHg and 90?C, is required to synthesize 23.0g CH3OH?
Pressure = 754 mmHg = 0.992 atm
Temperature = 90°C = 363 Kelvin
mass of CH3OH produced = 23.0 grams
Step 3: Calculate moles of CH3OH
Moles CH3OH = mass CH3OH / Molar mass CH3OH
Moles CH3OH = 23.0 grams / 32.04 g/mol
Moles CH3OH = 0.718 moles
Step 4: Calculate moles of H2
For 1 mol of CO consumed, we need 2 moles of H2 to produce 1 mol of CH3OH
For 0.718 moles CH3OH produced, we have 2*0.718 moles =1.436 moles of H2 and 0.718 moles of CO
Step 5: Calculate volume of H2
p*V = n*R*T
with p = the pressure = 0.992 atm
with V the volume = TO BE DETERMINED
with n = the number of moles = 1.436 moles H2
with R = the gasconstant = 0.08206 L*atm/ K*mol
with T = the temperature = 363 Kelvin
V = (n*R*T)/p
V = (1.436*0.08206*363)/0.992
V = 43.12 L
Step 6: Calculate volume of CO
p*V = n*R*T
with p = the pressure = 0.992 atm
with V the volume = TO BE DETERMINED
with n = the number of moles = 0.718 moles CO
with R = the gasconstant = 0.08206 L*atm/ K*mol
with T = the temperature = 363 Kelvin
V = (n*R*T)/p
V = (0.718*0.08206*363)/0.992
V = 21.56 L
A: There is 43.12 Liter of H2 needed
B: There is 21.56 liter of CO needed
