30g Mg and 30g `O_(2)` are reacted and the residual mixture contains:

Options

A. `C_(3)H_(6)`
B. `C_(2)H_(4)`
C. `CH_(4)`
D. `C_(6)H_(6)`

Answer

To solve the problem of what remains after reacting 30 g of magnesium (Mg) with 30 g of oxygen (O₂), we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between magnesium and oxygen 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

### Step 3: Calculate the number of moles of each reactant
- Moles of Mg:
\[ \text{Moles of Mg} = \frac{30 \text{ g}}{24 \text{ g/mol}} = 1.25 \text{ moles} \]

- Moles of O₂:
\[ \text{Moles of O}_2 = \frac{30 \text{ g}}{32 \text{ g/mol}} = 0.9375 \text{ moles} \]

### Step 4: Determine the limiting reagent
From the balanced equation, 2 moles of Mg react with 1 mole of O₂. Therefore, the required moles of O₂ for 1.25 moles of Mg can be calculated as:
\[ \text{Required O}_2 = \frac{1.25 \text{ moles Mg}}{2} = 0.625 \text{ moles O}_2 \]
Since we have 0.9375 moles of O₂ available, magnesium (Mg) is the limiting reagent.

### Step 5: Calculate the amount of product formed
Using the limiting reagent (Mg), we can find out how much magnesium oxide (MgO) is produced. According to the balanced equation, 2 moles of Mg produce 2 moles of MgO. Therefore, 1.25 moles of Mg will produce:
\[ \text{Moles of MgO} = 1.25 \text{ moles} \]

Now, calculate the mass of MgO produced:
\[ \text{Molar mass of MgO} = 24 \text{ g/mol (Mg)} + 16 \text{ g/mol (O)} = 40 \text{ g/mol} \]
\[ \text{Mass of MgO} = 1.25 \text{ moles} \times 40 \text{ g/mol} = 50 \text{ g} \]

### Step 6: Calculate the amount of excess reactant left
We initially had 30 g of O₂. The amount of O₂ that reacted can be calculated based on the moles of Mg used:
\[ \text{Moles of O}_2 \text{ reacted} = \frac{1.25 \text{ moles Mg}}{2} = 0.625 \text{ moles O}_2 \]
Now, convert this back to grams:
\[ \text{Mass of O}_2 \text{ reacted} = 0.625 \text{ moles} \times 32 \text{ g/mol} = 20 \text{ g} \]

Now, calculate the unreacted O₂:
\[ \text{Unreacted O}_2 = 30 \text{ g} - 20 \text{ g} = 10 \text{ g} \]

### Final Result
The residual mixture contains:
- 50 g of magnesium oxide (MgO)
- 10 g of unreacted oxygen (O₂)