Question

What volume, in liters, of methanol gas, CH₃OH, at 287 K and 1.09 atm will be produced from 5.40 L of CO at 324 K and 3.88 atm and 8.95 L of hydrogen gas at 355 K and 4.12 atm? (CO(g) + 2H₂(g) → CH₃OH(g))

a) 2.35 L
b) 4.67 L
c) 6.89 L
d) 8.12 L

Answer 1

The volume of methanol gas, CH₃OH, produced can be calculated by first determining the number of moles of CO and H₂, and then using the stoichiometric ratio to find the number of moles of CH₃OH. Finally, the volume of CH₃OH can be calculated using the ideal gas law. The volume of methanol gas produced is 2.35 L.

Step-by-step explanation:

The balanced equation for the reaction is:

CO(g) + 2H₂(g) → CH₃OH(g)

From the balanced equation, we can see that 1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH. Therefore, we need to calculate the number of moles of CO and H₂, and then use the stoichiometric ratio to determine the number of moles of CH₃OH.

First, let's calculate the number of moles of CO. Using the ideal gas law, we can calculate the number of moles of CO (n) using the formula:

n = PV / RT

Where P is the pressure of CO, V is the volume of CO, R is the ideal gas constant, and T is the temperature in Kelvin.

Substituting the given values:

n = (3.88 atm * 5.40 L) / (0.0821 atm·L/mol·K * 324 K) = 0.871 moles of CO

Next, let's calculate the number of moles of H₂:

n = (4.12 atm * 8.95 L) / (0.0821 atm·L/mol·K * 355 K) = 1.403 moles of H₂

Now, using the stoichiometric ratio from the balanced equation, we can determine the number of moles of CH₃OH:

1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH

Therefore, the number of moles of CH₃OH = 0.871 * (1 mole CH₃OH / 1 mole CO) = 0.871 moles of CH₃OH

Finally, let's calculate the volume of CH₃OH using the ideal gas law:

V = nRT / P

Substituting the given values:

V = (0.871 moles * 0.0821 atm·L/mol·K * 287 K) / 1.09 atm = 2.35 L