Question #578de

1 Answer
anor277
Jun 12, 2017

We would generate #2xx-128*kJ.......#

Explanation:

We has..........

#"Moles of dihydrogen"# #=# #(8.08*g)/(2.02*g*mol^-1)=4.0*mol#

#CO(g) + 2H_2(g) rarr CH_3OH(l)# #;DeltaH_"rxn"=-128*kJ#.

Enthalpy terms are written per mole of reaction as written. We reduce 4 moles of carbon monoxide, and should therefore generate #2xx-128*kJ=-256*kJ#.

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