Question

Determine how many kilojoules are released when 73.9 g of methanol reacts via the following reaction:2CH3OH(l)+3O2(g)→2CO2(g)+4H2O(l),ΔH=−726kJ

Answer 1

Step-by-step explanation

Given

Mass of methanol = 73.9 g

`2CH_3OH_((l))+3O_(2(g))\rightarrow2CO_(2(g))+4H_2O_((l))\text{ }\Delta H=-726kJ`

Solution

Step 1: Calculate the energy released per mole of methanol

Methanol has a coefficent of 2, meaning there are 2 moles.

`\begin{gathered} k\text{ = }\frac{-726kJ}{2\text{ moles}} \\ \\ k\text{ = -363 kJ/mol} \end{gathered}`

Step 2: calculate the number of moles of methanol

n = m/M where n is the moles, m is the mass and M is the molar mass

n = 73.9g/32,04 g/mol

n = 2.31 mol

Step 3: Calculate the enthalpy change

`\begin{gathered} \delta H=nk \\ \text{ = 2.31 mol x \lparen-363kJ\rparen} \\ \text{ = -837.26 kJ} \end{gathered}`

Answer

The energy (in KJ) of methanol released = -837.26 kJ