posted by Sarah
Problem#8: Metabolism of alcohol (ethanol, C2H5OH) by the body takes place according to the reaction:
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
ΔHo = -1367 kJ
Determine the number of “food calories” (Cal) supplied by the alcohol in a 12-oz (355-mL) bottle of “Molson Ice” beer that contains 4.50% alcohol by mass. Assume the density of the beer to be 1.00 g/mL. (1 Cal = 4.18 kJ; MM of C2H5OH = 46.1 g/mol)
4.50% by mass means 4.50 g EtOH/100 g solution. The density is 1.00 g/mL; therefore, that becomes 4.50 g EtOH/100 mL. How much alcohol is contained in 355 mL of the beer?
4.50 g EtOH x 355 mL/100 mL = 15.975 grams EtOH.
You get 1367 kJ for each 46 g EtOH according to the equation; therefore, there are
1367 kJ x 15.975/46 g = 474.73 kJ
Convert 474.73 kJ to kcal and that will be the same as the Cal (food energy).
Check my work. Check my thinking.