Question

Some drag-racing vehicles burn methanol, as presented by the following equation. What the mass of methanol that burns to produce an enthalpy change of `-9.00 xx 10^4` kJ?

Answer 1

Approx. `4.5*kg` of alcohol..............

Explanation:

Well, one way that is often useful conceptually is to regard energy, `Delta`, as a PRODUCT in the exothermic chemical reaction (as indeed it is!). And thus we can write:

`H_3COH + 3/2O_2 rarr CO_2(g) + 2H_2O+Delta`

EQUIVALENTLY..........

`H_3COH + 3/2O_2 rarr CO_2(g) + 2H_2O+637.9*kJ`

(I halved the equation to make the arithmetic easier! Of course I also had to halve the energy output given my approach.)

And we want an enthalpy change of `-9xx10^4*kJ`. We know that the combustion of 1 mol of reaction as written will yield `637.9*kJ`.

And thus the quotient, `(9xx10^4*kJ)/(637.9*kJ*mol^-1)=141.1*mol`, and when we say `"mol"` here we mean `"moles of reaction as written"`.

And since for each mole of reaction, there is `1*mol` methanol involved, the required mass of methanol is the product............

`141.9*molxx32.04*g*mol^-1=4520*g`