Answer
a) $Cl_2$ is the limiting reactant.
b) $m(Al_2Cl_6) = 5.073g$
c) $m(Al) = 1.674g$
Work Step by Step
Let's consider the reaction that takes place:
$2Al + 3Cl_2 \rightarrow Al_2Cl_6$
Since we are given the mass of both reagents, first we need to determine which one is the limiting reagent. We can calculate the amounts of both of reagents:
$n(Al) = \frac{m(Al)}{A(Al)} = \frac{2.7g}{27\frac{g}{mol}} = 0.1mol$
$n(Cl_2) = \frac{m(Cl_2)}{M(Cl_2)} = \frac{4.05g}{71\frac{g}{mol}} = 0.057mol$
a) Observing stoichiometric coefficients, we can conclude that the amount of aluminum required for the reaction is $\frac{2}{3}$ times the amount of chlorine, which means that for the reaction of $0.057mol$ of $Cl_2$, $0.038mol$ of $Al$ is required. Therefore, chlorine is the limiting reagent.
b) The amount of $Al_2Cl_6$ produced in this reaction is three times less than the amount of chlorine, hence the amount of the obtained $Al_2Cl_6$ is equal to $0.019mol$, in theory. That corresponds to the following mass:
$m(Al_2Cl_6) = n(Al_2Cl_6)\times M(Al_2Cl_6) = 0.019mol \times 267\frac{g}{mol} = 5.073g$
c) Since $0.038mol$ of $Al$ is required for the reaction, the excess amount of aluminum is equal to $0.1mol - 0.038mol = 0.062mol$, which corresponds to the following mass:
$m(Al) = n(Al)\times A(Al) = 0.062mol \times 27\frac{g}{mol} = 1.674g$
