Calculating CH3OH Yield: Limiting Reactions of CO and H2 in Chemical Reaction

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Discussion Overview

The discussion revolves around calculating the yield of methanol (CH3OH) produced from a reaction between carbon monoxide (CO) and hydrogen gas (H2). The focus is on determining the limiting reactant based on the given masses of the reactants and the stoichiometric relationships in the chemical equation.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant states the chemical reaction for the production of CH3OH and provides the initial masses of H2 and CO.
  • Another participant notes the stoichiometric ratio of the reactants, indicating that 1 mole of CO reacts with 2 moles of H2 to produce 1 mole of CH3OH, and translates this into mass terms.
  • A third participant emphasizes the need to identify which reactant will limit the reaction based on the provided masses and stoichiometric requirements.
  • Some participants express reluctance to provide direct answers without the original poster demonstrating effort in solving the problem.

Areas of Agreement / Disagreement

There is no consensus on how to proceed with the calculation, as some participants are willing to guide the original poster while others insist on the need for the poster to attempt the problem independently.

Contextual Notes

The discussion lacks specific calculations or assumptions regarding the ideal gas behavior of the reactants and does not resolve how to approach the limiting reactant determination.

Who May Find This Useful

Students or individuals interested in stoichiometry, chemical reactions, and yield calculations in chemistry.

starsun
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CH3OH is produced via the reaction CO(g) + 2 H2(g) -> CH3OH

A mixture of 1.20 g H2 and 7.45g CO are allowed to react.

What is the yield of CH3OH? [Assume theoretical yield in g is what is wanted here]
 
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one mole of CO reacts with two moles of H2 to give one mole of CH3OH.
Can you take it from here.
 
nope, sorry
 
According to the reaction, there's a 1:2 ratio between the reactants. Which means you need exactly 2 moles of hydrogen gas to react with 1 mole of carbon monoxide (translated to mass terms, this means you need 4g of hydrogen gas and 28 g of CO to form 32 g of methanol). How much do you have and how much of what is going to limit the reaction to a value less than the maximum possible yield if both reactants were in stoichiometric proportions? If you can answer these questions, you will be through.

Cheers
Vivek
 
starsun said:
nope, sorry
Well you'd better give it a shot, if you want to be helped. We can not help you, unless you show that you are making an effort to solve this by yourself, and have given it your best shot.

There are rules on this forum : we don't do homework for people.
 

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