This question concerns the stoichiometry of gases.

Question

Methanol, CH3OH, is also known as wood alcohol. It is manufactured by reacting carbon monoxide and hydrogen gases at high temperatures and extreme pressure. Assume 666.0 m3 CO and 1210.0 m3 H2 go into the reaction.

a. Which reactant is present in excess?

b. How much of the excess reactant remains after the reaction?

c. What volume of methanol vapor is produced, assuming the same conditions?

Answer 1

Carbon monoxide = CO

Hydrogen gas = H₂

CO(g) + 2H₂(g) ==> CH₃OH(g) .. balanced equation

Which reactant is in excess: an easy way to determine excess and limiting reactant is to simply divide mols of each reactant or volume of gas, by the corresponding coefficient in the balanced equation. Whichever value is less will represent the limiting reactant.

For CO(g): 666 m³ (÷1->666)

For H₂(g): 1210 m³ (÷2->605)

Since 605 is less than 666 H₂ is limiting and CO is in excess

How much excess (CO) remains after the reaction

1210 m³ H₂ x 1 m³ CO / 2 m³ H₂ = 605 m³ CO used

666 m³ CO - 605 m³ = 61 m³ CO remaining after reaction

Volume of CH₃OH(g) produced

1210 m³ H₂ x 1 m³ CH₃OH / 2 m³ H₂ = 605 m³ CH₃OH produced