# Question #7c266

May 1, 2017

It will react with 191.7 gram ${O}_{2}$ and produce 175.8 gram $C {O}_{2}$ and 107.9 gram ${O}_{2}$.

#### Explanation:

To do a calculation on a reaction, we first determine the mol ratio in which the reactants reacts and the products are formed. This ratio is already given in the equation by the coefficients.

$\textcolor{red}{\textcolor{w h i t e}{a a a} 1 \textcolor{w h i t e}{a a a a a a a} : \textcolor{w h i t e}{a a a} 3 \textcolor{w h i t e}{a a a} : \textcolor{w h i t e}{a a a a} 2 \textcolor{w h i t e}{a a a a} : \textcolor{w h i t e}{a a a} 3}$
${C}_{2} {H}_{5} O H \left(g\right) + 3 {O}_{2} \left(g\right) \to 2 C {O}_{2} \left(g\right) + 3 {H}_{2} O \left(g\right)$

To find out the mass of both the products, we use the law of conservation of mass, that states that mass cannot be created nor destroyed.

We have given that we burn 92 gram of ${C}_{2} {H}_{5} O H \left(g\right)$. Since we know the mol ratio of the reaction, we have to convert this mass into the number of moles of ${C}_{2} {H}_{5} O H \left(g\right)$. To do this, we use the molar mass of ${C}_{2} {H}_{5} O H \left(g\right)$.

The molar mass of ${C}_{2} {H}_{5} O H \left(g\right)$ can be calculated using the average masses of the atoms in ${C}_{2} {H}_{5} O H \left(g\right)$. Therefore we have

2 times a C-atom$\implies 2 \times 12.01 = 24.02$ g/mol
6 times a H-atom$\implies 6 \times 1.008 = 6.048$ g/mol
1 time an O-atom $\implies 1 \times 16.00 = 16.00$ g/mol
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a} \frac{\textcolor{w h i t e}{a a a a a a a a a a a}}{\textcolor{w h i t e}{a}} +$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a} 46.068$ gram/mol

$\frac{92}{46.068} = 1.997 \ldots$ mol ${C}_{2} {H}_{5} O H \left(g\right)$

Note: We also find the molar mass for the 3 other compounds, but the calculations aren't shown here since they are all similar.

Now we can use the mol ratio to calculate the number of moles that is reacted or is produced with 1.997 mol ${C}_{2} {H}_{5} O H \left(g\right)$.
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
${O}_{2}$ calculations

${C}_{2} {H}_{5} O H \left(g\right) + 3 {O}_{2} \left(g\right)$
$\textcolor{red}{\textcolor{w h i t e}{a a a} 1 \textcolor{w h i t e}{a a a a a a a} : \textcolor{w h i t e}{a a a} 3}$
$\textcolor{w h i t e}{a a a a a a a a a a a} \cancel{\text{\}}$
$\textcolor{w h i t e}{a a a} 1.997 \textcolor{w h i t e}{a a a a a a a} \textcolor{b l u e}{5.991 \text{ mol}}$

5.991 $\times$ molar mass = $5.991 \times 32.00 = 191.7$ gram ${O}_{2}$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
$C {O}_{2}$ calculations

${C}_{2} {H}_{5} O H \left(g\right) \to 2 C {O}_{2} \left(g\right)$
$\textcolor{red}{\textcolor{w h i t e}{a a a} 1 \textcolor{w h i t e}{a a a a a a a} : \textcolor{w h i t e}{a a a} 2}$
$\textcolor{w h i t e}{a a a a a a a a a a a} \cancel{\text{\}}$
$\textcolor{w h i t e}{a a a} 1.997 \textcolor{w h i t e}{a a a a a a a} \textcolor{b l u e}{3.994 \text{ mol}}$

3.994 $\times$ molar mass = $3.994 \times 44.01 = 175.8$ gram $C {O}_{2}$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
${H}_{2} O$ calculations

${C}_{2} {H}_{5} O H \left(g\right) \to 3 {H}_{2} O \left(g\right)$
$\textcolor{red}{\textcolor{w h i t e}{a a a} 1 \textcolor{w h i t e}{a a a a a a a} : \textcolor{w h i t e}{a a a} 3}$
$\textcolor{w h i t e}{a a a a a a a a a a a} \cancel{\text{\}}$
$\textcolor{w h i t e}{a a a} 1.997 \textcolor{w h i t e}{a a a a a a a} \textcolor{b l u e}{5.991 \text{ mol}}$

5.991 $\times$ molar mass = $5.991 \times 18.016 = 107.9$ gram ${H}_{2} O$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
The 3 reaction schemes showed above are just to visualise what we are doing with the mol ratio. These scheme's do not represent a chemical reaction. The calculations are more easily done when writing everything under the total equation (the one you provided) instead of first making the separate schemes, but this is just to show how it works..