Which of the following statements `is//are` correct"
The following reaction occurs:
`2Al + 3MnO overset(Delta)rarr Al_(2) O_(3) + 3 Mn`.
`108.0 g` of `Al` and `213.0 g` of `MnO` was heated to initiate the reaction. `(Mw "of" MnO = 71`, atomic weight of `Al = 13)`

Options

A. `Al` is present in excess
B. `MnO` is present is excess.
C. `54.0 g` of `Al` is required
D. `159.0 g` of `MnO` is in excess.

Explanation

To solve the problem step by step, let's analyze the reaction and calculate the moles of the reactants involved.

### Step 1: Write down the balanced chemical equation
The balanced equation is:
\[ 2 \text{Al} + 3 \text{MnO} \rightarrow \text{Al}_2\text{O}_3 + 3 \text{Mn} \]

### Step 2: Calculate the moles of Aluminum (Al)
Given:
- Mass of Al = 108.0 g
- Atomic weight of Al = 27 g/mol

Using the formula for moles:
\[ \text{Moles of Al} = \frac{\text{mass}}{\text{molar mass}} = \frac{108.0 \, \text{g}}{27 \, \text{g/mol}} = 4 \, \text{moles} \]

### Step 3: Calculate the moles of Manganese Oxide (MnO)
Given:
- Mass of MnO = 213.0 g
- Molar mass of MnO = 71 g/mol

Using the formula for moles:
\[ \text{Moles of MnO} = \frac{\text{mass}}{\text{molar mass}} = \frac{213.0 \, \text{g}}{71 \, \text{g/mol}} \approx 3 \, \text{moles} \]

### Step 4: Determine the limiting reactant
From the balanced equation, we know:
- 2 moles of Al react with 3 moles of MnO.

Now, we can find the stoichiometric ratio:
- For 4 moles of Al, we would need:
\[ \text{Moles of MnO required} = \frac{3}{2} \times 4 = 6 \, \text{moles} \]

Since we only have 3 moles of MnO, MnO is the limiting reactant, and Al is in excess.

### Step 5: Calculate the amount of Al required for the reaction
From the stoichiometry:
- 3 moles of MnO require 2 moles of Al.

Thus, for 3 moles of MnO:
\[ \text{Moles of Al required} = \frac{2}{3} \times 3 = 2 \, \text{moles} \]

Now, calculate the mass of Al required:
\[ \text{Mass of Al required} = \text{moles} \times \text{molar mass} = 2 \, \text{moles} \times 27 \, \text{g/mol} = 54 \, \text{g} \]

### Step 6: Calculate the excess amount of Al
Total mass of Al available = 108.0 g
Mass of Al required = 54.0 g

Excess mass of Al:
\[ \text{Excess Al} = 108.0 \, \text{g} - 54.0 \, \text{g} = 54.0 \, \text{g} \]

### Conclusion
- Al is in excess, and the excess amount is 54.0 g.
- MnO is the limiting reactant.

### Summary of Correct Statements
1. Aluminum (Al) is in excess.
2. 54 g of Al is required for the reaction.
3. MnO is the limiting reactant.