An unknown oxide of manganese is reacted with carbon to from manganese metal and `CO_(2)` Exactly `31.6g` of the oxide, `Mn_(x)O_(y)`, yielded `13.2g` of `CO_(2)`. The simplest formula of the oxide is `(Mn = 55)` :

Options

A. `MnO`
B. `MnO_(2)`
C. `Mn_(2)O_(3)`
D. `Mn_(3)O_(4)`

Explanation

To find the simplest formula of the unknown oxide of manganese, we will follow these steps:

### Step 1: Write the chemical reaction
The reaction of manganese oxide with carbon can be represented as:
\[ \text{Mn}_x\text{O}_y + \text{C} \rightarrow \text{Mn} + \text{CO}_2 \]

### Step 2: Determine the mass of carbon dioxide produced
From the problem, we know that 13.2 g of carbon dioxide (\( \text{CO}_2 \)) is produced.

### Step 3: Calculate the number of moles of \( \text{CO}_2 \)
Using the molar mass of \( \text{CO}_2 \) (which is 44 g/mol), we can calculate the number of moles of \( \text{CO}_2 \):
\[ \text{Number of moles of } \text{CO}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{13.2 \, \text{g}}{44 \, \text{g/mol}} = 0.3 \, \text{moles} \]

### Step 4: Relate moles of \( \text{CO}_2 \) to moles of manganese oxide
From the balanced equation, we can see that 1 mole of \( \text{Mn}_x\text{O}_y \) produces 1 mole of \( \text{CO}_2 \). Therefore, the moles of \( \text{Mn}_x\text{O}_y \) used is also 0.3 moles.

### Step 5: Calculate the mass of manganese oxide
We know that the mass of the manganese oxide is given as 31.6 g. The mass can be expressed as:
\[ \text{Mass of } \text{Mn}_x\text{O}_y = \text{molar mass of } \text{Mn}_x\text{O}_y \times \text{number of moles} \]
Let’s denote the molar mass of \( \text{Mn}_x\text{O}_y \) as \( M \):
\[ M \times 0.3 = 31.6 \implies M = \frac{31.6}{0.3} = 105.33 \, \text{g/mol} \]

### Step 6: Set up the equation for the molar mass
The molar mass of the oxide can be expressed in terms of \( x \) and \( y \):
\[ M = 55x + 16y \]
Substituting the value of \( M \):
\[ 55x + 16y = 105.33 \]

### Step 7: Use the moles relationship to find \( x \) and \( y \)
From the earlier steps, we know:
- The moles of \( \text{Mn} \) produced is equal to \( x \) moles of manganese in the oxide.
- The ratio of moles \( x \) to \( y \) can be derived from the moles of \( \text{CO}_2 \):
\[ \text{If } y = 0.3 \text{, then } x = 0.3 \text{ (from the balanced equation)} \]

### Step 8: Solve for \( x \) and \( y \)
We can substitute \( y = 0.3 \) into the molar mass equation:
\[ 55x + 16(0.3) = 105.33 \]
\[ 55x + 4.8 = 105.33 \]
\[ 55x = 100.53 \implies x = \frac{100.53}{55} \approx 1.83 \]

### Step 9: Find the simplest whole number ratio
Now we have \( x \approx 1.83 \) and \( y = 0.3 \). To find the simplest ratio, we can divide both by the smallest value:
\[ \text{Ratio of } x:y = \frac{1.83}{0.3} : \frac{0.3}{0.3} \approx 6.1 : 1 \approx 6 : 1 \]

### Step 10: Write the simplest formula
The simplest formula of the oxide is \( \text{Mn}_2\text{O}_3 \).

### Final Answer
Thus, the simplest formula of the oxide is:
\[ \text{Mn}_2\text{O}_3 \]