The composition of residual mixture will be if 30 g of Mg combines with 30 g of oxygen

Options

A. `40g MgO+20g, O_(2)`
B. `45 g MgO+15gO_(2)`
C. `50gMgO+10gO_(2)`
D. `60gMgO` only

Explanation

To find the composition of the residual mixture when 30 g of magnesium (Mg) combines 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: Determine the molar masses
- Molar mass of magnesium (Mg) = 24 g/mol
- Molar mass of oxygen (O2) = 32 g/mol (since O = 16 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:
\[ \text{Moles of O2} = \frac{30 \text{ g}}{32 \text{ g/mol}} = 0.9375 \text{ moles} \]

### Step 4: Determine the limiting reagent
From the balanced equation, we see that 2 moles of Mg react with 1 mole of O2. Therefore, we need:
- For 1.25 moles of Mg, the required moles of O2 = \( \frac{1.25}{2} = 0.625 \) moles of O2.

Since we have 0.9375 moles of O2 available, magnesium (Mg) is the limiting reagent.

### Step 5: Calculate the amount of MgO produced
According to the balanced equation, 2 moles of Mg produce 2 moles of MgO. Therefore, 1.25 moles of Mg will produce:
\[ 1.25 \text{ moles of Mg} \rightarrow 1.25 \text{ moles of MgO} \]

### Step 6: Calculate the mass of MgO produced
The molar mass of MgO = 24 g/mol (Mg) + 16 g/mol (O) = 40 g/mol.
Thus, the mass of MgO produced is:
\[ \text{Mass of MgO} = 1.25 \text{ moles} \times 40 \text{ g/mol} = 50 \text{ g} \]

### Step 7: Calculate the residual mass of reactants
- Initial mass of Mg = 30 g
- Initial mass of O2 = 30 g
- Total initial mass = 30 g + 30 g = 60 g

Mass of MgO produced = 50 g. Therefore, the residual mass (unreacted oxygen) is:
\[ \text{Residual mass} = \text{Total initial mass} - \text{Mass of MgO} = 60 \text{ g} - 50 \text{ g} = 10 \text{ g} \]

### Step 8: Determine the amount of unreacted oxygen
Since we started with 30 g of O2 and used only 0.625 moles (which is equivalent to 20 g of O2), the unreacted O2 is:
\[ \text{Unreacted O2} = 30 \text{ g} - 20 \text{ g} = 10 \text{ g} \]

### Final Composition of Residual Mixture
The residual mixture consists of:
- 10 g of unreacted O2
- 0 g of unreacted Mg (since all Mg is consumed)
- 50 g of MgO produced

### Summary
The composition of the residual mixture is:
- 10 g of O2
- 50 g of MgO