# For the reaction of 2"Al"(s) with 3"Cl"_2(g) to form 2"AlCl"_3(s), if "20 atoms Al" reacts with "45 molecules" of "Cl"_2, how many atoms of "Al" are needed to react with the excess "Cl"_2?

Sep 3, 2017

As you have, the balanced reaction was:

${\text{2Al"(s) + 3"Cl"_2(g) -> 2"AlCl}}_{3} \left(s\right)$

This can be done by looking at the stoichiometric coefficients for the reactants. Since $\text{Al}$ is $2 : 3$ with ${\text{Cl}}_{2}$, there must be $\text{2 atoms Al}$ for every ${\text{3 molecules Cl}}_{2}$.

That means that for every $\text{20 atoms Al}$, you expect to use up...

20 cancel"atoms Al" xx ("3 molecules Cl"_2)/(2 cancel"atoms Al")

$= {\text{30 molecules Cl}}_{2}$

But you have ${\text{45 molecules Cl}}_{2}$ available... so ${\text{Cl}}_{2}$ is in excess by

overbrace("45 molecules Cl"_2)^"Available" - overbrace("30 molecules Cl"_2)^"Needed" = overbrace(ul"15 molecules")^"Excess",

and $\text{Al}$ is the limiting reactant. Knowing the amount of ${\text{Cl}}_{2}$ in excess, you will need:

15 cancel("molecules Cl"_2) xx "2 atoms Al"/(3 cancel("molecules Cl"_2))

$=$ $\underline{\textcolor{b l u e}{\text{10 more atoms Al}}}$

to use up all $\text{15 molecules}$ of leftover ${\text{Cl}}_{2}$.