1.0 g of Mg is burnt with 0.28 g of `O_2` in a closed vessel . Which reactant is left in excess and how much ?

Options

A. Mg , 5.8 g
B. Mg , 0.58 g
C. `O_2 , 0.24` g
D. `O_2` , 2.4 g

Answer

To solve the problem of which reactant is left in excess and how much, we will follow these steps:

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

### Step 2: Determine the molar masses
- Molar mass of Mg = 24 g/mol
- Molar mass of O₂ = 32 g/mol

### Step 3: Calculate the moles of each reactant
- Moles of Mg:
\[ \text{Moles of Mg} = \frac{1.0 \text{ g}}{24 \text{ g/mol}} = 0.04167 \text{ mol} \]

- Moles of O₂:
\[ \text{Moles of O}_2 = \frac{0.28 \text{ g}}{32 \text{ g/mol}} = 0.00875 \text{ mol} \]

### Step 4: Determine the stoichiometric ratio
From the balanced equation, 2 moles of Mg react with 1 mole of O₂. Therefore:
- 1 mole of O₂ reacts with 2 moles of Mg.
- This means that 0.00875 moles of O₂ would react with:
\[ 0.00875 \text{ mol O}_2 \times 2 = 0.0175 \text{ mol Mg} \]

### Step 5: Compare the available moles
We have:
- Available moles of Mg = 0.04167 mol
- Required moles of Mg for the available O₂ = 0.0175 mol

Since 0.04167 mol of Mg is greater than 0.0175 mol, magnesium is in excess.

### Step 6: Calculate the excess amount of Mg
To find out how much Mg is left unreacted, we subtract the amount of Mg that reacted from the initial amount:
- Amount of Mg that reacted with 0.00875 moles of O₂ = 0.0175 mol (as calculated)
- Mass of Mg that reacted:
\[ \text{Mass of Mg that reacted} = 0.0175 \text{ mol} \times 24 \text{ g/mol} = 0.42 \text{ g} \]

- Remaining Mg:
\[ \text{Remaining Mg} = 1.0 \text{ g} - 0.42 \text{ g} = 0.58 \text{ g} \]

### Conclusion
The reactant that is left in excess is magnesium (Mg), and the amount left is 0.58 g.

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