91Ó°ÊÓ

We are given the equation and we have to find the mole-mole factors.

In the given equation, there are ways to find mole factors by considering two compounds every time.

Consider,$AlandC{l}_2$

$2$mole Al=data-custom-editor="chemistry" $3$mole data-custom-editor="chemistry" ${\mathrm{Cl}}_2$

Now considerdata-custom-editor="chemistry" ${\mathrm{Cl}}_2\mathrm{and}{\mathrm{AlCl}}_3$

$3$mole$C{l}_2=2$ moledata-custom-editor="chemistry" ${\mathrm{AlCl}}_3$

which gives the data-custom-editor="chemistry" $\frac{3\mathrm{mole}{\mathrm{Cl}}_2}{2\mathrm{mole}{\mathrm{AlCl}}_3}$mole-mole factor.

Again, consider data-custom-editor="chemistry" $\mathrm{Al}\mathrm{and}{\mathrm{AlCl}}_3$

data-custom-editor="chemistry" $2$mole Al=data-custom-editor="chemistry" $2$mole data-custom-editor="chemistry" ${\mathrm{AlCl}}_3$

which gives the data-custom-editor="chemistry" $\frac{2\mathrm{mole}{\mathrm{AlCl}}_3}{2\mathrm{mole}\mathrm{Al}}$mole-mole factor.

Hence, these are the mole-mole factors.

We are given the equation and we have to find the mole-mole factors.

In the given equation, there are ways to find mole factors by considering two compounds every time.

Consider HCl and ${\mathrm{O}}_2$

$4$mole HCl = $1$mole${\mathrm{O}}_2$

which gives the$\frac{4\mathrm{mole}\mathrm{HCl}}{1\mathrm{mole}{\mathrm{O}}_2}$ mole-mole factor.

Now consider,${\mathrm{O}}_2\mathrm{and}{\mathrm{Cl}}_2$

$1$mole${\mathrm{O}}_2=2$ mole${\mathrm{Cl}}_2$

which gives the$\frac{1\mathrm{mole}{\mathrm{O}}_2}{2\mathrm{mole}{\mathrm{Cl}}_2}$ mole-mole factor.

Again, consider HCl and ${\mathrm{Cl}}_2$

$4$mole HCl= $2$mole${\mathrm{Cl}}_2$

which gives the$\frac{2\mathrm{mole}{\mathrm{Cl}}_2}{4\mathrm{mole}\mathrm{HCl}}$ mole-mole factor.

Hence, these are the mole-mole factors.