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? (At.wt: Mg=24, O = 16)

Options

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

Answer

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

### Step 1: Write the balanced chemical equation
The reaction of magnesium with oxygen can be represented as:
\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \]

### Step 2: Calculate the number of moles of magnesium (Mg)
Given:
- Mass of Mg = 1.0 g
- Atomic weight of Mg = 24 g/mol

Using the formula for moles:
\[ \text{Number of moles of Mg} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1.0 \, \text{g}}{24 \, \text{g/mol}} = 0.04167 \, \text{mol} \]

### Step 3: Calculate the number of moles of oxygen (O₂)
Given:
- Mass of O₂ = 0.56 g
- Molar mass of O₂ = 32 g/mol (since O has an atomic weight of 16, O₂ = 16 x 2)

Using the formula for moles:
\[ \text{Number of moles of O}_2 = \frac{0.56 \, \text{g}}{32 \, \text{g/mol}} = 0.0175 \, \text{mol} \]

### Step 4: Determine the limiting reactant
From the balanced equation, we see that 1 mole of O₂ reacts with 2 moles of Mg. Therefore, the moles of Mg required for the available moles of O₂ can be calculated as follows:
\[ \text{Moles of Mg required} = 2 \times \text{Moles of O}_2 = 2 \times 0.0175 \, \text{mol} = 0.035 \, \text{mol} \]

We have 0.04167 moles of Mg available, which is more than the 0.035 moles required. Thus, O₂ is the limiting reactant.

### Step 5: Calculate the remaining moles of magnesium
Now, we can find out how much magnesium is left after the reaction:
\[ \text{Remaining moles of Mg} = \text{Initial moles of Mg} - \text{Moles of Mg consumed} \]
\[ \text{Remaining moles of Mg} = 0.04167 \, \text{mol} - 0.035 \, \text{mol} = 0.00667 \, \text{mol} \]

### Step 6: Convert remaining moles of magnesium to grams
To find the mass of the remaining magnesium:
\[ \text{Mass of remaining Mg} = \text{Remaining moles} \times \text{Molar mass of Mg} = 0.00667 \, \text{mol} \times 24 \, \text{g/mol} = 0.16 \, \text{g} \]

### Conclusion
The reactant left in excess is magnesium (Mg), and the amount left is approximately 0.16 g.

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