Calculate heat of combustion of cell for the following reaction CH11.66N0.2O0.27+1.28O2->Co2+0.1N2+0.83H2o. If the heat of combustion of substrate is 104kJ/g, yH=56KJ/g cells

Answer 1

I'm guessing the reaction is supposed to be something like this....

CH_1.66N_0.2O_0.27 + 1.28O₂ ==> CO₂ + 0.1N₂ + 0.83H₂O ... balanced equation for combustion reaction

∆Hrxn = heat of combustion = 104 kJ/g substrate

∆Hrxn = heat of combustion = 56 kJ/g cells

1 mol substrate = 311 g (C15H25N3O4)

From here, I'm still not certain what your question is. Are you asking for heat of combustion per cell or per g of cells? The way you phrase the question "Calculate heat of combustion of cell..." isn't really clear. Sorry I can't be of any further help at this point. Maybe another tutor will be able to better understand what it is you are asking.