Question

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`?

Answer 1

As you have, the balanced reaction was:

`"2Al"(s) + 3"Cl"_2(g) -> 2"AlCl"_3(s)`

This can be done by looking at the stoichiometric coefficients for the reactants. Since `"Al"` is `2:3` with `"Cl"_2`, there must be `"2 atoms Al"` for every `"3 molecules Cl"_2`.

That means that for every `"20 atoms Al"`, you expect to use up...

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

`= "30 molecules Cl"_2`

But you have `"45 molecules Cl"_2` available... so `"Cl"_2` is in excess by

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

and `"Al"` is the limiting reactant. Knowing the amount of `"Cl"_2` in excess, you will need:

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

`=` `ulcolor(blue)"10 more atoms Al"`

to use up all `"15 molecules"` of leftover `"Cl"_2`.