Q153CP

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?

Step-by-Step Solution

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Answer

The mass can be found as 466.2 g Mn₂O₃, 76.1 g MnO.
The ratio can be found as 5.52/1 Mn³⁺/Mn²⁺ ratio.

1Step 1: 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,

$M o l e s o f M n_{2} O_{3} \times \frac{1 m o l M n_{2} O_{3}}{157.87 g M n_{2} O_{3}} = \frac{x}{157.87} m o l M n_{2} O_{3} M o l e s o f M n o = y g M n O \times \frac{1 m o l M n O}{70.94 g M n O} = \frac{y}{70.94} m o l M n O$

The moles of Mn and O can be,

$M o l e s M n (f r o m M n_{2} O_{3}) = \frac{x}{157.87} m o l M n_{2} O_{3} \times \frac{2 m o l M n}{1 m o l M n_{2} O_{3}} M o l e s M n (f r o m M n_{2} O_{3}) = 0.0127 x m o l M n M o l e s M n (f r o m M n_{2} O_{3}) = \frac{x}{157.87} m o l M n_{2} O_{3} \times \frac{3 m o l M n}{1 m o l M n_{2} O_{3}} M o l e s M n (f r o m M n_{2} O_{3}) = 0.0190 x m o l M n$

Then,

$M o l e s M n (f r o m M n O) = \frac{y}{70.94} m o l M n O \times \frac{1 m o l M n}{1 m o l M n O} M o l e s M n (f r o m M n O) = 0.0141 y m o l M n M o l e s M n (f r o m M n O) = \frac{y}{70.94} m o l M n O \times \frac{1 m o l M n}{1 m o l M n O} M o l e s M n (f r o m M n O) = 0.0141 y m o l O$

On isolating the first equation,

x+y=542.3 g

Y=542.3 g-x

On solving g or x,

$\begin{array}{rcl} x & = & 466.2 g M n_{2} O_{3} \\ y & = & 542.3 g - x = 542.3 g - 466.2 g \\ = & 76.1 g M n O \end{array}$

2Step 2: Finding the ratio Mn 3+ /Mn 2+ in the ore

The moles can be denoted as,

$M o l e s o f M n^{3 +} = 466.2 g M n_{2} O_{3} \times \frac{1 m o l M n_{2} O_{3}}{157.87 g M n_{2} O_{3}} \times \frac{2 m o l M n^{3 +}}{1 m o l M n_{2} O_{3}} M o l e s o f M n^{3 +} = 5.91 m o l M o l e s o f M n^{2 +} = 76.1 g M n O \times \frac{1 m o l M n O}{157.87 g M n O} \times \frac{2 m o l M n^{2 +}}{1 m o l M n O} M o l e s o f M n^{2 +} = 1.07 m o l \frac{M n^{3 +}}{M n^{2 +}} = \frac{5.91 m o l M n^{3 +}}{1.07 m o l M n^{2 +}} = \frac{5.52}{1.00}$

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