Question

5.46 aluminium chloride, alcl₃(s), is made by reacting scrap aluminium with chlorine: 2 al(s) + 3 cl₂(g) → 2 alcl₃(s) suppose a reaction vessel has 2.70 g of aluminium and 4.05 g of chlorine. (a) which reactant limits the amount of reaction that can happen? (b) what is the maximum mass of alcl₃(s) that can be produced? (c) what is the mass of excess reactant that would remain after reaction has gone to completion? (d) set up an amounts table for this reaction mixture.

Explanation:

Step1: Calculate moles of reactants

The molar mass of $Al$ is $M_{Al}=26.98\ g/mol$, and the molar mass of $Cl_2$ is $M_{Cl_2}=2\times35.45 = 70.90\ g/mol$.
The number of moles of $Al$, $n_{Al}=\frac{m_{Al}}{M_{Al}}=\frac{2.70\ g}{26.98\ g/mol}\approx0.100\ mol$.
The number of moles of $Cl_2$, $n_{Cl_2}=\frac{m_{Cl_2}}{M_{Cl_2}}=\frac{4.05\ g}{70.90\ g/mol}\approx0.0571\ mol$.

Step2: Determine the limiting reactant

From the balanced - chemical equation $2Al(s)+3Cl_2(g)\to2AlCl_3(s)$, the mole ratio of $Al$ to $Cl_2$ is $\frac{n_{Al}}{n_{Cl_2}}=\frac{2}{3}$.
If $Al$ is the limiting reactant, the moles of $Cl_2$ required to react with $0.100\ mol$ of $Al$ is $n_{Cl_2, required}=\frac{3}{2}n_{Al}=\frac{3}{2}\times0.100\ mol = 0.150\ mol$. But we only have $0.0571\ mol$ of $Cl_2$.
If $Cl_2$ is the limiting reactant, the moles of $Al$ required to react with $0.0571\ mol$ of $Cl_2$ is $n_{Al, required}=\frac{2}{3}n_{Cl_2}=\frac{2}{3}\times0.0571\ mol\approx0.0381\ mol$. Since $0.0381\ mol<0.100\ mol$, $Cl_2$ is the limiting reactant.

Step3: Calculate the maximum mass of $AlCl_3$

From the balanced - chemical equation, the mole ratio of $Cl_2$ to $AlCl_3$ is $\frac{n_{AlCl_3}}{n_{Cl_2}}=\frac{2}{3}$.
The moles of $AlCl_3$ produced, $n_{AlCl_3}=\frac{2}{3}n_{Cl_2}=\frac{2}{3}\times0.0571\ mol\approx0.0381\ mol$.
The molar mass of $AlCl_3$ is $M_{AlCl_3}=26.98 + 3\times35.45=133.33\ g/mol$.
The mass of $AlCl_3$ produced, $m_{AlCl_3}=n_{AlCl_3}\times M_{AlCl_3}=0.0381\ mol\times133.33\ g/mol\approx5.08\ g$.

Step4: Calculate the mass of the excess reactant

The moles of $Al$ that react with $0.0571\ mol$ of $Cl_2$ is $n_{Al, reacted}=\frac{2}{3}n_{Cl_2}\approx0.0381\ mol$.
The moles of $Al$ remaining, $n_{Al, remaining}=n_{Al}-n_{Al, reacted}=0.100\ mol - 0.0381\ mol = 0.0619\ mol$.
The mass of $Al$ remaining, $m_{Al, remaining}=n_{Al, remaining}\times M_{Al}=0.0619\ mol\times26.98\ g/mol\approx1.67\ g$.

Step5: Set up the amounts table

Substance	Initial moles	Change in moles	Final moles
$Cl_2$	$0.0571$	$- 0.0571$	$0$
$AlCl_3$	$0$	$0.0381$	$0.0381$

Answer:

(a) Chlorine ($Cl_2$) is the limiting reactant.
(b) The maximum mass of $AlCl_3$ is approximately $5.08\ g$.
(c) The mass of the excess reactant ($Al$) remaining is approximately $1.67\ g$.
(d)

Substance	Initial moles	Change in moles	Final moles
$Cl_2$	$0.0571$	$- 0.0571$	$0$
$AlCl_3$	$0$	$0.0381$	$0.0381$