# Question 578de

Jun 12, 2017

We would generate $2 \times - 128 \cdot k J \ldots \ldots .$

#### Explanation:

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

$\text{Moles of dihydrogen}$ $=$ $\frac{8.08 \cdot g}{2.02 \cdot g \cdot m o {l}^{-} 1} = 4.0 \cdot m o l$

$C O \left(g\right) + 2 {H}_{2} \left(g\right) \rightarrow C {H}_{3} O H \left(l\right)$ ;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 $2 \times - 128 \cdot k J = - 256 \cdot k J$.