Answer:
The balanced equation for the decomposition of mercury(II) oxide (HgO) is:
\[ 2HgO \rightarrow 2Hg + O_2 \]
Given that 1.252 g of mercury(II) oxide (HgO) decomposes, we need to determine the number of moles of mercury(II) oxide and then use stoichiometry to find the number of moles of diatomic oxygen produced.
1. Calculate the molar mass of HgO:
- Hg: 200.59 g/mol
- O: 16.00 g/mol
- Molar mass of HgO: \(200.59 + 16.00 = 216.59\) g/mol
2. Determine the number of moles of HgO:
\[ \text{Moles of HgO} = \frac{\text{Mass of HgO}}{\text{Molar mass of HgO}} \]
\[ \text{Moles of HgO} = \frac{1.252 \text{ g}}{216.59 \text{ g/mol}} \approx 0.00577 \text{ mol} \]
3. Using the stoichiometry of the reaction, we find the moles of oxygen produced:
- From the balanced equation, 2 moles of HgO produce 1 mole of \(O_2\).
- So, 1 mole of HgO produces \( \frac{1}{2} \) moles of \(O_2\).
Thus, the number of moles of \(O_2\) produced is:
\[ \text{Moles of } O_2 = \frac{1}{2} \times 0.00577 \text{ mol} = 0.002885 \text{ mol} \]
4. Convert the number of moles of \(O_2\) to mass in milligrams:
\[ \text{Mass of } O_2 = \text{Moles of } O_2 \times \text{Molar mass of } O_2 \times 1000 \]
\[ \text{Molar mass of } O_2 = 2 \times 16.00 \text{ g/mol} = 32.00 \text{ g/mol} \]
\[ \text{Mass of } O_2 = 0.002885 \text{ mol} \times 32.00 \text{ g/mol} \times 1000 = 92.32 \text{ mg} \]
Therefore, the information requested is:
- The balanced equation for the decomposition of mercury(II) oxide is \( 2HgO \rightarrow 2Hg + O_2 \).
- The mass of diatomic oxygen formed by the decomposition of 1.252 g of mercury(II) oxide is 92.32 milligrams.
