Answer
$Al \space and \space Cl_2:$ $\frac{2 \space moles \space Al}{3 \space moles \space Cl_2}$ and $\frac{3 \space moles \space Cl_2}{2 \space moles \space Al}$
$Al \space and \space AlCl_3:$ $\frac{2 \space moles \space Al}{2 \space moles \space AlCl_3}$ and $\frac{2 \space moles \space AlCl_3}{2 \space moles \space Al}$
$Cl_2 \space and \space AlCl_3: $ $\frac{3 \space mole \space Cl_2}{2 \space moles \space AlCl_3}$ and $\frac{2 \space moles \space AlCl_3}{3 \space mole \space Cl_2}$
Work Step by Step
1. Use the balance coefficients to write the mole-mole equalities:
2 moles $Al$ = 3 moles $Cl_2$
2 moles $Al$ = 2 moles $AlCl_3$
3 moles $Cl_2$ = 2 moles $AlCl_3$
2. Use these equalities to write the conversion factors:
$\frac{2 \space moles \space Al}{3 \space moles \space Cl_2}$ and $\frac{3 \space moles \space Cl_2}{2 \space moles \space Al}$
$\frac{2 \space moles \space Al}{2 \space moles \space AlCl_3}$ and $\frac{2 \space moles \space AlCl_3}{2 \space moles \space Al}$
$\frac{3 \space mole \space Cl_2}{2 \space moles \space AlCl_3}$ and $\frac{2 \space moles \space AlCl_3}{3 \space mole \space Cl_2}$
