How many grams of # CH_3OH# can be made from 23.12 g of #CO# and 8.323 of #H_2#?

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Answer 1 · 2017-02-11T22:14:42

We need (i) a stoichiometric equation........

#CO + 2H_2 rarr CH_3OH#

And (ii) equivalent quantities of carbon monoxide and dihydrogen.

#"Moles of carbon monoxide"# #=# #(23.12*g)/(28.01*g*mol^-1)=0.825*mol#.

#"Moles of dihydrogen"# #=# #(8.323*g)/(2.016*g*mol^-1)=2.016*mol#.

Clearly, the dihydrogen is in stoichiometric excess, and at most we can make #0.825*mol# of methyl alcohol.

And thus, maximum quantity of methyl alcohol,

#=0.825*molxx32.04*g*mol^-1=26.4*g#.

The mixture of carbon monoxide and dihydrogen is called syngas. This is usually used directly without conversion to methyl alcohol.