How much heat is produced by combustion 125 g of methanol under standard state condaitions?

For the reaction,

2CH₃OH + 3O₂ → 2CO₂ + 4 H₂O

For the above reaction,

the change in enthalphy is calculated as

Δ Hrxn = Δ H°f (products) - Δ H°f (reactants)

In case the compound is in its standard state, enthalphy of formation is zero

Hence,

for the above reaction,

ΔHrxn = (2 * Δ H° (CO₂) + 4 * Δ H° (H₂O)) - [ (2 * Δ H°CH₃OH) + (3 * Δ H° O₂) ]

Δ H° (CO₂) = - 393.5kJ / mol

Δ H° (H₂O) = - 241.8 kJ / mol

Δ H°CH₃OH = - 239.2kJ / mol

Δ H° O₂ = 0

putting the corresponding values,

ΔHrxn = (2 * - 393.5kJ / mol + 4 * - 241.8kJ / mol) - [ (2 * -239.2kJ / mol) + (3 * 0)

ΔHrxn = - 1275.8 kJ / mol

Moles of methanol,

Moles is denoted by given mass divided by the molecular mass,

Hence,

n = w / m

n = moles,

w = given mass,

m = molecular mass.

From the question,

w = 125 g

as we know,

m = 32 g / mol

n = 125 g / 32 g / mol = 3.906 mol

From, the reaction, 2 mol produces - 1275.8 kJ / mol heat,

Now using unitary method,

1 mol produces = - 1275.8 kJ / mol / 2 heat,

3.906 mol produces = - 1275.8 kJ / mol / 2 * 3.906 heat

3.906 mol produces = 249.7 kJ