Question

When methanol, CH 3 OH , is burned in the presence of oxygen gas, O 2 , a large amount of heat energy is released. For this reason, it is often used as a fuel in high performance racing cars. The combustion of methanol has the balanced, thermochemical equation CH 3 OH ( g ) + 3 2 O 2 ( g ) ⟶ CO 2 ( g ) + 2 H 2 O ( l ) Δ H = − 764 kJ How much methanol, in grams, must be burned to produce 581 kJ of heat?

Answer 1

Answer: The mass of methanol that must be burned is 24.34 grams

Explanation:

We are given:

Amount of heat produced = 581 kJ

For the given chemical equation:

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H=-764kJ$

By Stoichiometry of the reaction:

When 764 kJ of heat is produced, the amount of methanol reacted is 1 mole

So, when 581 kJ of heat will be produced, the amount of methanol reacted will be = $\frac{1}{764}\times 581=0.7605mol$

To calculate mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of methanol = 0.7605 moles

Molar mass of methanol = 32 g/mol

Putting values in above equation, we get:

$0.7605mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.7605mol\times 32g/mol)=24.34g$

Hence, the mass of methanol that must be burned is 24.34 grams