# 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?

Apr 13, 2017

#### Answer:

Approx. $4.5 \cdot k g$ 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:

${H}_{3} C O H + \frac{3}{2} {O}_{2} \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O + \Delta$

EQUIVALENTLY..........

${H}_{3} C O H + \frac{3}{2} {O}_{2} \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O + 637.9 \cdot k J$

(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 $- 9 \times {10}^{4} \cdot k J$. We know that the combustion of 1 mol of reaction as written will yield $637.9 \cdot k J$.

And thus the quotient, $\frac{9 \times {10}^{4} \cdot k J}{637.9 \cdot k J \cdot m o {l}^{-} 1} = 141.1 \cdot m o l$, and when we say $\text{mol}$ here we mean $\text{moles of reaction as written}$.

And since for each mole of reaction, there is $1 \cdot m o l$ methanol involved, the required mass of methanol is the product............

$141.9 \cdot m o l \times 32.04 \cdot g \cdot m o {l}^{-} 1 = 4520 \cdot g$