# COMPLETE COMBUSTION of C_(14)H_(16)O_3 to CO_2 and H_2O. You have 315.0000 grams of C_(14)H_(16)O_3. What is the limiting reagent? How many grams of CO_2 are generated? How many grams of H_2O are generated?

May 7, 2018

Given that we are told only the mass of ${C}_{14} {H}_{16} {O}_{3}$, we can assume that it undergoes combustion in excess oxygen, so ${C}_{14} {H}_{16} {O}_{3}$ is the limiting reagent.

Mass of $C {O}_{2} = 836.5$ $g$
Mass of ${H}_{2} O = 195.6$ $g$

#### Explanation:

First step is always to create a balanced chemical reaction:

${C}_{14} {H}_{16} {O}_{3} + \frac{33}{2}$ ${O}_{2} \to 14$ $C {O}_{2} + 8$ ${H}_{2} O$

If you're uncomfortable with a fractional coefficient the whole equation could be multiplied by 2, but it's not necessary.

Now, the molar mass of ${C}_{14} {H}_{16} {O}_{3}$ is $232$ $g m o {l}^{-} 1$.

The number of moles of ${C}_{14} {H}_{16} {O}_{3}$ is $n = \frac{m}{M} = \frac{315}{232} = 1.358$ $m o l$

Each mole of ${C}_{14} {H}_{16} {O}_{3}$ produces 14 mole of $C {O}_{2}$, and $C {O}_{2}$ has a molar mass of $44$ $g m o {l}^{-} 1$, so the mass of $C {O}_{2}$ produced is:

$1.358 \times 14 \times 44 = 836.5$ $g$ of $C {O}_{2}$.

By the same reasoning for ${H}_{2} O$ with its molar mass of $18$ $g m o {l}^{-} 1$:

$1.358 \times 8 \times 18 = 195.6$ $g$ of ${H}_{2} O$.