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#?

Question

Caleb Adams · Accepted Answer

Just as you have, the thoughtful response 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#.

Henry Antrim · Answer

undefined To find the number of atoms of Al needed to react with the excess Cl2, first, determine the limiting reactant:

1. Calculate the number of moles of Al and Cl2 using the given amounts and Avogadro's number.
2. Use the stoichiometry of the reaction to determine which reactant limits the amount of product formed.
3. Once you find the limiting reactant, use the stoichiometry of the reaction to calculate the amount of excess reactant left.
4. Finally, convert the amount of excess reactant to the number of atoms of Al needed.

Given the balanced equation:
$$ 2 \text{Al}(s) + 3 \text{Cl}_2(g) \rightarrow 2 \text{AlCl}_3(s) $$

1. Calculate the number of moles of Al:
$$ \text{20 atoms Al} \times \frac{1 \text{ mol Al}}{6.022 \times 10^{23} \text{ atoms Al}} = \text{moles of Al} $$

2. Calculate the number of moles of Cl2:
$$ \text{45 molecules Cl}_2 \times \frac{1 \text{ mol Cl}_2}{6.022 \times 10^{23} \text{ molecules Cl}_2} = \text{moles of Cl}_2 $$

3. Determine the limiting reactant by comparing the moles of Al and Cl2 and use stoichiometry to find the excess reactant.

4. Once the limiting reactant is identified, use stoichiometry to find the number of excess atoms of Al.

5. The excess atoms of Al can be calculated by subtracting the atoms used in the limiting reactant reaction from the total atoms initially available.