Methanol Production Process and Calculation

Aug 14, 2024

Methanol Production from Carbon Monoxide and Hydrogen

Reaction Overview

  • Methanol is produced by the reaction of carbon monoxide (CO) and hydrogen (H2).
  • Given quantities for reaction:
    • Carbon monoxide: 356 grams
    • Hydrogen: 65 grams
  • Task: Determine the mass of methanol produced and the mass of excess reactant remaining after the reaction.

Balancing the Chemical Equation

  • Ensure the chemical equation is balanced:
    • Carbon: 1 atom on both sides
    • Oxygen: 1 atom on both sides
    • Hydrogen: 4 atoms on both sides (2x2 on the left, 3+1 on the right)
  • Conclusion: The equation is balanced.

Determining Limiting and Excess Reactants

  • Limiting Reactant: The reactant that determines the amount of product formed.
  • Excess Reactant: The reactant that remains after the reaction.

Calculating Moles

  • Carbon Monoxide (CO):

    • Molecular weight = 28 (Carbon = 12, Oxygen = 16)
    • Moles of CO = 356 grams / 28 = 12.7 moles

  • Hydrogen (H2):

    • Molecular weight = 2 (2 atoms of Hydrogen)
    • Moles of H2 = 65 grams / 2 = 32.5 moles

Stoichiometric Ratio

  • Reaction requires 2 moles of H2 per mole of CO.
  • Given:
    • 32.5 moles H2 and 12.7 moles CO
    • Ratio given: 2.56 moles H2 per mole CO
  • Conclusion:
    • Excess Reactant: Hydrogen
    • Limiting Reactant: Carbon Monoxide

Calculating Methanol Production

  • From stoichiometry, 1 mole of CO produces 1 mole of methanol (CH3OH).
  • Moles of methanol produced = Moles of limiting reactant (CO) = 12.7 moles
  • Molecular weight of methanol = 32 (C=12, H=3, O=16, additional H=1)
  • Mass of methanol produced = 12.7 moles x 32 g/mole = 406 grams

Calculating Excess Reactant Remaining

  • Total mass of reactants = 356 grams (CO) + 65 grams (H2) = 421 grams
  • Mass of product (methanol) = 406 grams
  • Mass of excess reactant = 421 grams - 406 grams = 15 grams

Verification through Stoichiometry

  • 12.7 moles CO needs 25.4 moles H2 (2:1 ratio)
  • H2 used = 25.4 moles x 2 g/mole = 50.8 grams
  • Initial H2 = 65 grams, H2 remaining = 65 - 50.8 = 14.2 grams

Summary

  • Mass of Methanol Produced: 406 grams
  • Excess Reactant Remaining: Approximately 15 grams of hydrogen
  • Minor discrepancies in rounding observed.