Let's say we have an exothermic, entropy decreasing reaction where $\Delta$H is negative and $\Delta$S is also negative. $$ \Delta G = \Delta H - T \Delta S $$ $$H_2 + \frac{1}{2} O_2 \rightarrow H_2 O$$
The -T$\Delta$S term will be positive meaning energy will be lost because the products "contain less heat" (decrease in entropy) than the reactants. I guess where I'm confused is why can't this lost heat be used as work? Why is it "lost"?
"Contain less heat" is indeed confusing because it's nonsensical; it seems to refer to the debunked theory of caloric. Materials don't "contain heat."
This reaction destroys a lot of entropy locally* by removing a gas from the universe. One might multiply this entropy change by a temperature to obtain a term with units of energy, but I'm not sure it's useful to think of this term as something you could have extracted. What type of machine would you construct? Upon completion of the reaction, the gas is gone; it's now incorporated into the liquid water.
*The energy given off by the formation of water molecules goes into heating the universe, which increases the surrounding entropy. The net global entropy change is positive under conditions where the reaction is spontaneous.
It may be useful to consider an even simpler reaction: liquid water and ice at 0°C (with the surroundings also at 0°C). There's no driving force to move from 20%/80% water/ice to 50%/50% water ice, for instance, and you can't connect an engine that would extract any energy from that transition. The $\Delta G$ for that compositional change is zero. Yet $\Delta H$ and $T\Delta S$ are both nonzero (and offset each other exactly; indeed, this is what defines an equilibrium phase transition temperature).
