91Ó°ÊÓ

The first equation can be denoted as,

x+y=542.3 g

The moles can be denoted as,

$$\begin{gathered} MolesofM{n}_2{O}_3×\frac{1molM{n}_2{O}_3}{157.87gM{n}_2{O}_3}=\frac{x}{157.87}molM{n}_2{O}_3 \\ MolesofMno=ygMnO×\frac{1molMnO}{70.94gMnO}=\frac{y}{70.94}molMnO \end{gathered}$$

The moles of Mn and O can be,

$$\begin{gathered} MolesMn\left(fromM{n}_2{O}_3\right)=\frac{x}{157.87}molM{n}_2{O}_3×\frac{2molMn}{1molM{n}_2{O}_3} \\ MolesMn\left(fromM{n}_2{O}_3\right)=0.0127xmolMn \\ MolesMn\left(fromM{n}_2{O}_3\right)=\frac{x}{157.87}molM{n}_2{O}_3×\frac{3molMn}{1molM{n}_2{O}_3} \\ MolesMn\left(fromM{n}_2{O}_3\right)=0.0190xmolMn \end{gathered}$$

Then,

$$\begin{gathered} MolesMn\left(fromMnO\right)=\frac{y}{70.94}molMnO×\frac{1molMn}{1molMnO} \\ MolesMn\left(fromMnO\right)=0.0141ymolMn \\ MolesMn\left(fromMnO\right)=\frac{y}{70.94}molMnO×\frac{1molMn}{1molMnO} \\ MolesMn\left(fromMnO\right)=0.0141ymolO \end{gathered}$$

On isolating the first equation,

x+y=542.3 g

Y=542.3 g-x

On solving g or x,

$\begin{array}{rcl}x& =& 466.2gM{n}_2{O}_3\\ y& =& 542.3g-x=542.3g-466.2g\\ & =& 76.1gMnO\end{array}$

The moles can be denoted as,

$$\begin{gathered} MolesofM{n}^{3+}=466.2gM{n}_2{O}_3×\frac{1molM{n}_2{O}_3}{157.87gM{n}_2{O}_3}×\frac{2molM{n}^{3+}}{1molM{n}_2{O}_3} \\ MolesofM{n}^{3+}=5.91mol \\ MolesofM{n}^{2+}=76.1gMnO×\frac{1molMnO}{157.87gMnO}×\frac{2molM{n}^{2+}}{1molMnO} \\ MolesofM{n}^{2+}=1.07mol \\ \frac{M{n}^{3+}}{M{n}^{2+}}=\frac{5.91molM{n}^{3+}}{1.07molM{n}^{2+}}=\frac{5.52}{1.00} \end{gathered}$$