Methanol (CH3OH) can be produced through the reaction of carbon monoxide and hydrogen in the presence of a catalyst. If 75.0 grams of carbon monoxide reacts to produce 68.4 grams of methanol, what is the percent yield of the reaction?
Please show all work.
Thank you soooo much!
Please show all work.
Thank you soooo much!
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CO+ 2 H2 -----> CH3OH: 1 mole Co reacts to form 1 mole methanol
calc moles CO: 75 g CO/ 28 g/mole CO = ? moles CO
? moles CO x 32 g/mole CH3OH= theo. am't methanol
% yield= actual yield/theo. yield x 100
% Yield= 68.4/ theo. yield x 100
calc moles CO: 75 g CO/ 28 g/mole CO = ? moles CO
? moles CO x 32 g/mole CH3OH= theo. am't methanol
% yield= actual yield/theo. yield x 100
% Yield= 68.4/ theo. yield x 100