One mole of a hydrocarbon is combusted. The products obtained are cooled down to STP and occupy a volume of $89.6\ \mathrm l$. Oxygen required for combustion was $145.6\ \mathrm l$ at STP. Find the molecular formula of the hydrocarbon.
Let the hydrocarbon be $\ce{C}_x\ce{H}_y$. The combustion reaction can be written as: $$ \ce{C_xH_y + O2 -> H2O + CO2} $$ Keep in mind that this is not yet balanced.
We can now find out the amount of reactants and products. Since one mole of any gas occupies $22.4\ \mathrm l$ of space at STP, we can say that we have: $$ \frac{89.6\ \mathrm l}{22.4\ \mathrm l/\mathrm{mol}} = 4\ \mathrm{mol} $$ of products. As for the oxygen reacted: $$ \frac{145.6\ \mathrm l}{22.4\ \mathrm l/\mathrm{mol}} = 6.5\ \mathrm{mol} $$. We can now deduce that the hydrocarbon will react with $13\ce{O}_2$. Therefore, the equation becomes. $$ \ce{C_xH_y + 13O2 -> H2O + CO2} $$ But, There are $x$ $\ce{C}$s one the left hand side so the right hand side also has $x$ $\ce{C}$s: $$ \ce{C_xH_y + 13O2 -> H2O + xCO2} $$ There are $y$ $\ce{H}$'s on the left hand side. Same for the right hand-side: $$ \ce{C_xH_y + 13O2 -> y/2H2O + xCO2} $$ Since the amount of $\ce{O}$ has to balance as well, we have $26$ oxygens on the LHS and therefore $26$ on the RHS. Which means: $$ 26 = \frac{y}{2} + 2x \implies \frac{y}{2} = 26 - 2x $$ We have already figured out that $4\:\mathrm{mol}$ of products are obtained. i.e.: $$ 4 = \frac{y}{2} + x = 26 - 2x + x = 26 - x \implies x = 26 - 4 = 22 $$ But this would mean that $y$ is negative. Where did I go wrong?