# Question #27862

Apr 14, 2016

${\text{4.5 moles Cl}}_{2}$

#### Explanation:

You're dealing with a synthesis reaction in which aluminium metal, $\text{Al}$, reacts with chlorine gas, ${\text{Cl}}_{2}$, to form aluminium chloride, ${\text{AlCl}}_{3}$.

The balanced chemical equation that describes this reaction looks like this

$2 {\text{Al"_ ((s)) + color(red)(3)"Cl"_ (2(g)) -> color(blue)(2)"AlCl}}_{3 \left(s\right)}$

Notice that you have a $\textcolor{red}{3} : \textcolor{b l u e}{2}$ mole ratio between chlorine gas and aluminium chloride. This tells you that for every $\textcolor{red}{3}$ moles of chlorine gas that take part in the reaction, the reaction produces $\textcolor{b l u e}{2}$ moles of aluminium chloride.

In other words, the reaction will consume $\frac{\textcolor{red}{3}}{\textcolor{b l u e}{2}}$ times more moles of chlorine gas than the number of moles of aluminium chloride it will produce.

The problem tells you that the reaction produced $3$ moles of aluminium chloride. This means that the reaction must have consumed

$3 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles AlCl"_3))) * (color(red)(3)color(white)(a)"moles Cl"_2)/(color(blue)(2)color(red)(cancel(color(black)("moles AlCl"_3)))) = color(green)(|bar(ul(color(white)(a/a)"4.5 moles Cl}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

As a conclusion, the stoichiometric coefficients listed in the balanced chemical equation can be used as mole ratios between the chemical species that take part in a given reaction.