Answer
a. $ Al $ and $ Cl_2 $: $$ \frac{ 2 \space moles \space Al }{ 3 \space moles \space Cl_2 } \space and \space \frac{ 3 \space moles \space Cl_2 }{ 2 \space moles \space Al }$$
$ Al $ and $ AlCl_3 $: $$ \frac{ 2 \space moles \space Al }{ 2 \space moles \space AlCl_3 } \space and \space \frac{ 2 \space moles \space AlCl_3 }{ 2 \space moles \space Al }$$
$ Cl_2 $ and $ AlCl_3 $: $$ \frac{ 3 \space moles \space Cl_2 }{ 2 \space moles \space AlCl_3 } \space and \space \frac{ 2 \space moles \space AlCl_3 }{ 3 \space moles \space Cl_2 }$$
b. $ HCl $ and $ O_2 $: $$ \frac{ 4 \space moles \space HCl }{ 1 \space mole \space O_2 } \space and \space \frac{ 1 \space mole \space O_2 }{ 4 \space moles \space HCl }$$
$ HCl $ and $ Cl_2 $: $$ \frac{ 4 \space moles \space HCl }{ 2 \space moles \space Cl_2 } \space and \space \frac{ 2 \space moles \space Cl_2 }{ 4 \space moles \space HCl }$$
$ HCl $ and $ H_2O $: $$ \frac{ 4 \space moles \space HCl }{ 2 \space moles \space H_2O } \space and \space \frac{ 2 \space moles \space H_2O }{ 4 \space moles \space HCl }$$
$ O_2 $ and $ Cl_2 $: $$ \frac{ 1 \space mole \space O_2 }{ 2 \space moles \space Cl_2 } \space and \space \frac{ 2 \space moles \space Cl_2 }{ 1 \space mole \space O_2 }$$
$ O_2 $ and $ H_2O $: $$ \frac{ 1 \space mole \space O_2 }{ 2 \space moles \space H_2O } \space and \space \frac{ 2 \space moles \space H_2O }{ 1 \space mole \space O_2 }$$
$ Cl_2 $ and $ H_2O $: $$ \frac{ 2 \space moles \space Cl_2 }{ 2 \space moles \space H_2O } \space and \space \frac{ 2 \space moles \space H_2O }{ 2 \space moles \space Cl_2 }$$
Work Step by Step
1. Identify all pairs:
a. We can find the mole-mole factors between: $Al$ and $Cl_2$, $Al$ and $AlCl_3$, $Cl_2$ and $AlCl_3$ .
b. We can find the mole-mole factors between: $HCl$ and $O_2$, $HCl$ and $Cl_2$, $HCl$ and $H_2O$, $O_2$ and $Cl_2$, $O_2$ and $H_2O$, $Cl_2$ and $H_2O$.
2. Use the coefficients of the balanced equation to write the conversion factors.
a. $2Al(s) +3Cl_2(g) \longrightarrow 2AlCl_3(s)$
Thus, if we use 2 $Al$ moles, we will need 3 $Cl_2$ moles to react with it, and it will produce 2 $AlCl_3$ moles.
2 $Al$ moles = 3 $Cl_2$ moles
2 $Al$ moles = 2 $AlCl_3$ moles
3 $Cl_2$ moles = 2 $AlCl_3$ moles
b. $4HCl(g) + O_2(g) \longrightarrow 2Cl_2(g) + 2H_2O(g)$
Thus, if we use 4 $HCl$ moles, we will need 1 $O_2$ mole to react with it, and it will produce 2 $Cl_2$ moles and 2 $H_2O$ moles.
4 $HCl$ moles = 1 $O_2$ mole
4 $HCl$ moles = 2 $Cl_2$ moles
4 $HCl$ moles = 2 $H_2O$ moles
1 $O_2$ mole = 2 $Cl_2$ moles
1 $O_2$ mole = 2 $H_2O$ moles
2 $Cl_2$ moles = 2 $H_2O$ moles
