1g of Mg is burnt in a vessel containing 0.5 g of oxygen. The remaining unreacted is

Options

A. 0.25 g of Mg
B. 0.1 g of Mg
C. 0.1 g of `O_(2)`
D. 0.75 g of Mg

Answer

To solve the problem of determining the remaining unreacted magnesium after burning 1g of magnesium in a vessel containing 0.5g of oxygen, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between magnesium (Mg) and oxygen (O2) can be represented by the following balanced equation:
\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \]

### Step 2: Calculate the molar masses
- Molar mass of magnesium (Mg) = 24 g/mol
- Molar mass of oxygen (O2) = 32 g/mol (since O has a molar mass of 16 g/mol)

### Step 3: Calculate the number of moles of magnesium and oxygen
- For magnesium:
\[ \text{Number of moles of Mg} = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{24 \text{ g/mol}} \approx 0.04167 \text{ moles} \]

- For oxygen:
\[ \text{Number of moles of O}_2 = \frac{0.5 \text{ g}}{32 \text{ g/mol}} \approx 0.015625 \text{ moles} \]

### Step 4: Determine the limiting reagent
Using the stoichiometric coefficients from the balanced equation:
- From the equation, 1 mole of O2 reacts with 2 moles of Mg.
- Therefore, the number of moles of Mg required for 0.015625 moles of O2 is:
\[ \text{Moles of Mg required} = 2 \times 0.015625 \approx 0.03125 \text{ moles} \]

### Step 5: Compare the available moles
- Available moles of Mg = 0.04167 moles
- Required moles of Mg = 0.03125 moles

Since the required moles of Mg (0.03125) are less than the available moles (0.04167), oxygen (O2) is the limiting reagent.

### Step 6: Calculate the moles of magnesium that react
From the stoichiometry, if 0.015625 moles of O2 react, then:
\[ \text{Moles of Mg that react} = 2 \times 0.015625 = 0.03125 \text{ moles} \]

### Step 7: Calculate the remaining moles of magnesium
\[ \text{Remaining moles of Mg} = \text{Initial moles of Mg} - \text{Moles of Mg that react} = 0.04167 - 0.03125 = 0.01042 \text{ moles} \]

### Step 8: Convert the remaining moles of magnesium to grams
\[ \text{Mass of remaining Mg} = \text{Remaining moles} \times \text{Molar mass of Mg} = 0.01042 \text{ moles} \times 24 \text{ g/mol} \approx 0.25 \text{ g} \]

### Final Answer
The remaining unreacted magnesium is approximately **0.25 grams**.

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