1.0 g of magnesium is burnt with 0.56 g `O_(2)` in a closed vessel. Which reactant is left in excess and how much?

Options

A. (a)`Mg, 0.16 g`
B. (b)`O_(2), 0.16 g`
C. (c )`Mg, 0.44 g`
D. (d)`O_(2), 0.28 g`

Explanation

To determine which reactant is left in excess and how much, we need to follow these steps:

1. **Write the balanced chemical equation:**
\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \]

2. **Calculate the molar masses:**
- Molar mass of Mg = 24 g/mol
- Molar mass of O\(_2\) = 2 × 16 g/mol = 32 g/mol

3. **Determine the stoichiometric ratio:**
According to the balanced equation:
- 2 moles of Mg react with 1 mole of O\(_2\)
- In terms of mass: 48 g of Mg react with 32 g of O\(_2\)

4. **Calculate the amount of O\(_2\) required to react with 1 g of Mg:**
\[ \text{Mass of O}_2 \text{ required} = \left(\frac{32 \text{ g O}_2}{48 \text{ g Mg}}\right) \times 1 \text{ g Mg} = 0.67 \text{ g O}_2 \]

5. **Compare the available O\(_2\) with the required O\(_2\):**
- Available O\(_2\) = 0.56 g
- Required O\(_2\) = 0.67 g

Since 0.56 g of O\(_2\) is less than 0.67 g, O\(_2\) is the limiting reagent.

6. **Calculate the amount of Mg that reacts with 0.56 g of O\(_2\):**
\[ \text{Mass of Mg that reacts} = \left(\frac{48 \text{ g Mg}}{32 \text{ g O}_2}\right) \times 0.56 \text{ g O}_2 = 0.84 \text{ g Mg} \]

7. **Determine the amount of Mg left unreacted:**
\[ \text{Mass of Mg left} = 1 \text{ g Mg (initial)} - 0.84 \text{ g Mg (reacted)} = 0.16 \text{ g Mg} \]

Therefore, the reactant left in excess is magnesium (Mg), and the amount left is 0.16 g.