# Tag Info

Since the reaction is done in excess of air, you can assume complete combustion and accordingly you should break down the reaction to only this. Apparently the question demands an answer in the following form: $$\ce{C8H18 + 1.2a (O2 + 3.76 N2) -> b CO2 + c H2O + d O2 + e N2},$$ however, this is the same as $$\begin{multline} \ce{C8H18 + x O2 + 0.2x O2 + ... -2 Key things I did not consider properly: 120% of excess air, which means 0.2 of oxygen goes free, this changes my equation to:$$\ce{C8H18 + 1.2$a$ (O2 + 3.76 N2) -> b CO2 +c H2O + 0.2a O2 + e N2} So, \begin{align} \ce{C}:&& b &= 8\\ \ce{H}:&& 2c &= 18\\ \ce{O}:&& 2b + c + 2(0.2) &= 2\times1.2a \implies a = 12.5\\ \... 1 I'll continue after volume of \ce H_2O vapours = 0.5 ml Vapour pressure is equal to the partial pressure of water in gas phase. P_{\ce H_2O} = P_{total} × mole fraction of \ce H_2O Where P_{total} is 1 atm and mole fraction = 0.5/25 SoP_{\ce H_2O} = \frac{1}{50} atm = \frac{760}{50} mm Hg Now for the solution of non volatile solute $A$ ...