12 g of Mg was burnt in a closed vessel containing 32 g oxygen , which of the following is/are correct.

Options

A. (a)2 gm of Mg will be lift unburnt.
B. (b)0.75 gm-molecules of `O_(2)` will be left unreacted .
C. (c)20 gm of MgO will be formed.
D. (d)The mixture at the end will weight 44 g.

Answer

To solve the problem, we need to analyze the combustion of magnesium in the presence of oxygen and determine the correctness of the given statements based on the stoichiometry of the reaction.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between magnesium (Mg) and oxygen (O₂) can be represented as:
\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \]

2. **Calculate Moles of Reactants**:
- **Moles of Magnesium**:
Given mass of Mg = 12 g
Molar mass of Mg = 24 g/mol
\[ \text{Moles of Mg} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{12 \text{ g}}{24 \text{ g/mol}} = 0.5 \text{ moles} \]

- **Moles of Oxygen**:
Given mass of O₂ = 32 g
Molar mass of O₂ = 32 g/mol (16 g/mol for each O atom)
\[ \text{Moles of O}_2 = \frac{32 \text{ g}}{32 \text{ g/mol}} = 1 \text{ mole} \]

3. **Determine Limiting Reactant**:
According to the balanced equation, 2 moles of Mg react with 1 mole of O₂. Therefore, 0.5 moles of Mg will react with:
\[ \text{Required moles of O}_2 = \frac{0.5 \text{ moles Mg}}{2} = 0.25 \text{ moles O}_2 \]
Since we have 1 mole of O₂ available, magnesium is the limiting reactant.

4. **Calculate Remaining Reactants**:
- Moles of O₂ that remain unreacted:
\[ \text{Unreacted O}_2 = 1 \text{ mole (initial)} - 0.25 \text{ moles (reacted)} = 0.75 \text{ moles} \]

5. **Calculate Mass of MgO Formed**:
From the reaction, 0.5 moles of Mg will produce 0.5 moles of MgO. The molar mass of MgO is:
\[ \text{Molar mass of MgO} = 24 \text{ g/mol (Mg)} + 16 \text{ g/mol (O)} = 40 \text{ g/mol} \]
Therefore, the mass of MgO produced is:
\[ \text{Mass of MgO} = 0.5 \text{ moles} \times 40 \text{ g/mol} = 20 \text{ g} \]

6. **Calculate Total Mass of the Mixture**:
The total mass after the reaction will be the sum of the masses of the reactants:
\[ \text{Total mass} = \text{mass of Mg} + \text{mass of O}_2 = 12 \text{ g} + 32 \text{ g} = 44 \text{ g} \]

### Conclusion:
Now we can evaluate the statements:
- **Statement A**: 2 grams of magnesium will be left unburned. **(False)**
- **Statement B**: 0.75 molecules of oxygen will be left unreacted. **(False)**
- **Statement C**: 20 grams of MgO will be formed. **(True)**
- **Statement D**: The mixture at the end will weigh 44 grams. **(True)**

### Final Answer:
The correct statements are **C** and **D**.