Question

Manganese is a key component of extremely hard steel. The element occurs naturally in many oxides. A 542.3-g sample of a manganese oxide has an Mn/O ratio of 1.00/1.42 and consists of braunite (Mn₂O₃) and manganosite (MnO). (a) What masses of braunite and manganosite are in the ore? (b) What is the ratio Mn³⁺/Mn²⁺ in the ore?

Short answer

1. The mass can be found as466.2 g Mn₂O₃, 76.1 g MnO.
2. The ratio can be found as5.52/1 Mn³⁺/Mn²⁺ ratio.

Step-by-step solution

Finding the masses of braunite and manganosite

The first equation can be denoted as,

x+y=542.3 g

The moles can be denoted as,

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

The moles of Mn and O can be,

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

Then,

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

On isolating the first equation,

x+y=542.3 g

Y=542.3 g-x

On solving g or x,

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

Finding the ratio Mn3+/Mn2+ in the ore

The moles can be denoted as,

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