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?

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1 Answer
anor277
Apr 13, 2017

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#

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