Question

Question #578de

Answer 1

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`.