Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: jvr1581 on November 11, 2010, 11:17:28 PM
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I have the balanced equation...Methanol is synthesized from the combination of hydrogen and carbon monoxide
CO + H2O ---> CH3OH
Suppose 56.8 g of Carbon monoxide reacts with 8.6 g of hydrogen. What is the limiting reagent and theoretical yield?
This is what I have so far, but I'm not sure if it is correct:
56.8g CO x (1 mol/28.0105g CO) (1 mol CH3OH/1 mol CO) = 2.027 mol CH3OH
8.6g H x (1 mol/4.028g H) ( 1 mol CH3OH/2 mol H2) = 1.067 mol CH3OH
Are these calculations correct and is hydrogen the limiting reagent?
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I have the balanced equation...Methanol is synthesized from the combination of hydrogen and carbon monoxide
CO + H2O ---> CH3OH
CO + 2 H2 ---> CH3OH
You wrote the sentence correct but the reaction presented was incorrect.
Edit; 8.6g H x (1 mol/4.028g H) ( 1 mol CH3OH/2 mol H2) = 1.067 mol CH3OH
I don't understand why you go from H to H2 and why 1mol of H is 4.028? You are reacting hydrogen (H2) in the reaction, which is 2.014g/mol. The first step is to determine how many moles you have, so 8.6g H2 x 1mol/2.014g H2. Then the conversion factor mole over mole.
According to the way you've done it you have 8.6g hydrogen divided by (4.028g)x(2) = 8.05600g, which means there are four hydrogens molecules (H2) or eight individual atoms of hydrogen (H). The reaction says this is not correct, there is a total of 2 moles of hydrogen (H2). You should replace 4.028g with 2.014g.
Did I misunderstand something?
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Thanks, that is why I was asking. I was confused. You answered my question