Answer
See answer below.
Work Step by Step
Number of moles of the reactants:
Aluminum: $2.70\ g\div 26.983 g/mol=0.100\ mol$
Chlorine: $4.05\ g\div 70.906\ g/mol=0.057\ mol$
Ratio: $0.057/0.100=0.57$
Chlorine is the limiting reactant.
From stoichiometry:
$0.057\ mol\times 1_{AlCl_3}/3_{Cl_2} \times 133.34\ g/mol\ AlCl_3=2.54\ g\ AlCl_3$
Aluminum reacted:
$0.057\ mol\times 2_{Al}/3_{Cl_2} \times 26.982\ g/mol\ Al=1.03\ g\ Al$
Remaining: $2.70-1.03=1.67\ g\ Al$
$\begin{smallmatrix}
&2\ Al&3\ Cl_2&\rightarrow&2\ AlCl_3\\
Initial &0.100&0.057&&0\\
Reacted&-0.038&-0.057&&+0.038\\
Final&0.062&0&&0.038\\
\end{smallmatrix}$
