Question

What is the change in energy, in kJ, when 45.3 grams of methanol, CH3OH, combusts? 2\text{CH}_3\text{OH}\left(l\right) + 3\text{O}_2\left(g\right)\rightarrow2\text{CO}_2\left(g\right)+4\text{H}_2\text{O}\left(g\right)\hskip .5in \Delta\text{H}=-726\text{ kJ}2 CH 3 OH ( l ) + 3 O 2 ( g ) → 2 CO 2 ( g ) + 4 H 2 O ( g ) Δ H = − 726 kJ Group of answer choices -513 kJ +2,050 kJ -1,030 kJ -2,050 kJ +513 kJ

Answer 1

The change in energy when 45.3 grams of methanol combusts is approximately -1026 kJ.

Step-by-step explanation:

When 45.3 grams of methanol, CH3OH, combusts, we can calculate the change in energy by using the given enthalpy change, ΔH = -726 kJ. We need to convert grams of methanol to moles by using its molar mass, which is 32.04 g/mol.

First, find the number of moles: 45.3 g / 32.04 g/mol = 1.414 mol.

Since the enthalpy change is given per mole of methanol combusted, we can multiply the moles by the enthalpy change to calculate the change in energy: 1.414 mol × -726 kJ/mol = -1026 kJ. Therefore, the change in energy when 45.3 grams of methanol combusts is approximately -1026 kJ.