# How much heat is produced when 29.6 g of methanol react with 45.9 g of oxygen?

Nov 19, 2016

You need to supply the enthalpy of combustion of methanol.

#### Explanation:

Methanol combusts according to the reaction:

${H}_{3} C O H + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O + \Delta$

This reaction will be exothermic. Just how exothermic it will be is subject to measurement, which you have not quoted.

$\text{Moles of methanol} = \frac{29.6 \cdot g}{32.04 \cdot g \cdot m o {l}^{-} 1} = 0.924 \cdot m o l$.

$\text{Moles of dioxygen} = \frac{45.9 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} = 1.41 \cdot m o l$.

Given the molar quantities, there is sufficient dioxygen for complete combustion. However, we still need the molar enthalpy of combustion.