Methanol can be produced through the reaction of CO and H2 in the presence of a catalyst. CO(g) + 2 H2(g) catalyst −−−−−→ CH3OH(ℓ)?
If 71g of CO reacts to produce 64.2 g of
CH3OH, what is the percent yield of CH3OH?
Answer in units of %.
More Questions From This User See All
Answers & Comments
Verified answer
CO + 2 H2 → CH3OH
Supposing excess H2:
(71 g CO) / (28.0101 g CO/mol) x (1 mol CH3OH / 1 mol CO) x
(32.0419 g CH3OH/mol) = 81.2198 g CH3OH in theory
(64.2 g) / (81.2198 g) = 0.79045 = 79% yield CH3OH
You have to assume that H2 is present in excess in order to answer the question. But assuming that H2 is present in excess, then:
Calculate the theoretical yield:
71 g CO / 28.0 g/mol X (1 mol CH3OH / 1 mol CO) X 32.04 g/mol = 81 g/mol
% Yield = (actual yield / theoretical yield) X 100
% yield = (64.2 / 81) X 100 = 79%