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

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The discussion centers on the production of methanol (CH3OH) from the reaction of carbon monoxide (CO) and hydrogen gas (H2). The reaction requires a 1:2 molar ratio of CO to H2, meaning 1 mole of CO reacts with 2 moles of H2 to produce 1 mole of CH3OH. Given the amounts of 1.20 g of H2 and 7.45 g of CO, participants emphasize the need to determine which reactant is the limiting factor for the yield of methanol. Theoretical calculations indicate that 4 g of H2 and 28 g of CO are needed to produce 32 g of methanol. The discussion highlights the importance of understanding stoichiometry in chemical reactions and encourages participants to engage actively in problem-solving rather than relying on others to provide answers.
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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