# Tag Info

Let Hydrocarbon be of form $\ce{C_xH_y}$ Our combustion reaction becomes $\ce{C_xH_y}+(2x+\frac{y}{2})\ce{O_2}\implies x\ce{CO_2}+\frac{y}{2}\ce{H_2O}$ $1$ mole of hydrocarbon produce $x*44g$ of $\ce{CO_2}$ If $\ce{CO_2}$ produced is $33.01$, then $x=\frac{3}{4}$ Similarly $1$ mole of hydrocarbon produces $\frac{y}{2}$ moles of $\ce{H_2O}$ If $\ce{H_2O}$ ...