[jvr1581]

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?

[Fluorine]

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?

[jvr1581]

Thanks,  that is why I was asking.  I was confused.  You answered my question