Question 32eca

Feb 18, 2016

$\text{8 moles H"_2"O}$

Explanation:

The problem tells you that methyl alcohol, which I'll refer to as methanol from this point on, $\text{CH"_3"OH}$, reacts with oxygen gas ,${\text{O}}_{2}$, to form carbon dioxide, ${\text{CO}}_{2}$, and water, $\text{H"_2"O}$.

The first thing to do here is write a balanced chemical equation for this combustion reaction. Start with the unbalanced chemical equation which has methanol and oxygen as reactants and carbon dioxide and water as products

${\text{CH"_3"OH"_text((l]) + "O"_text(2(g]) -> "CO"_text(2(g]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

In order to have a balanced chemical equation, you must have equal numbers of atoms of each element on both sides of the equation.

Look at carbon. You have one carbon atom on the reactants' side and one on the products' side, so you don't need to make any changes.

Next, focus on hydrogen. You have $4$ atoms of hydrogen on the reactants' side and only $2$ on the products' side. To balance the hydrogen atoms, multiply the water by $2$

${\text{CH"_3"OH"_text((l]) + "O"_text(2(g]) -> "CO"_text(2(g]) + 2"H"_2"O}}_{\textrm{\left(l\right]}}$

Finally, look at the oxygen atoms. You have $3$ on the reactants' side and $4$ on the products' side. A useful trick to remember here is that when you're dealing with diatomic molecules like ${\text{O}}_{2}$, you can use fractional coefficients as a way to help the balancing process.

If you multiply the oxygen molecule by $\frac{3}{2}$, you will get a total of $4$ oxygen atoms on the reactants' side

${\text{CH"_3"OH"_text((l]) + 3/2"O"_text(2(g]) -> "CO"_text(2(g]) + 2"H"_2"O}}_{\textrm{\left(l\right]}}$

To get rid of the fractional coefficient, multiply everything by $2$. This will get you

$2 {\text{CH"_3"OH"_text((l]) + color(red)(3)"O"_text(2(g]) -> 2"CO"_text(2(g]) + color(blue)(4)"H"_2"O}}_{\textrm{\left(l\right]}}$

Now, you know that this reaction consumes $6$ moles of oxygen gas. In order to figure out how many moles of water will be produced, look at the $\textcolor{red}{3} : \textcolor{b l u e}{4}$ mole ratio that exists between the two compounds.

This mole ratio tells you that for every $\textcolor{red}{3}$ moles of oxygen gas that take part in the reaction, $\textcolor{b l u e}{4}$ moles of water are produced.

So, if $3$ moles of oxygen gas react, you can expect the reaction to produce $4$ moles of water. This means that when $6$ moles of oxygen react, you will get

6 color(red)(cancel(color(black)("moles O"_2))) * (color(blue)(4)color(white)(a)"moles H"_2"O")/(color(red)(3)color(red)(cancel(color(black)("moles O"_2)))) = color(green)("8 moles H"_2"O")#