Methanol CH3Oh can be produced by reacting carbon monoxide and hydrogen gas in the presence of a catalyst. If 75.00 grams of carbon monoxide reacts with 106.3 grams of hydrogen, 68.40 grams of methanol are produced. What is the percent yield of methanol?

Question

can someone help me set this problem up.

Answer 1

Which is the limiting reagent?
CO + 2H2 ==> CH3OH
mols CO = gram/molar mass = approx 3
mols H2 = approx 53
Convert mols CO to mols CH3OH using the coefficients in the balanced equation. That's 3 x (1 mol CH3OH/1 mol CO) = approx 3
Do the same for H2 to CH3OH. That's
53 mol H2 x (1 mol CH3OH/2 mol H2) = appxox 26
You note you have two values for mols CH3OH and both can't be right. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent responsible for that number is the limiting reagent. So all of the CO will be used, you will have an excess of H2 remaining unreacted, and you will produce approx 3 mols CH3OH.
Now convert 3 mols CH3OH to grams. g = mols x molar mass. This is the theoretical yield (that is 100% yield). The actual yield in the problem is given as 68.4g
%yield = (actual yield/theor yield)*100 = ?

Answer 2

what is the theoretical yield

Answer 3

Read my response. Next to last sentence.
"Now convert 3 mols CH3OH to grams. g = mols x molar mass. This is the theoretical yield (that is 100% yield)."