1. Balanced Chemical Equation:
The decomposition of mercury(II) oxide (HgO) is represented by the following equation:
2 HgO(s) → 2 Hg(l) + O₂(g)
2. Moles of HgO:
* Find 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
* Calculate moles of HgO:
* moles = mass / molar mass
* moles = 5.0 g / 216.59 g/mol = 0.0231 mol HgO
3. Moles of O₂:
* Use the mole ratio from the balanced equation: 2 moles HgO produce 1 mole O₂
* moles of O₂ = (0.0231 mol HgO) * (1 mol O₂ / 2 mol HgO) = 0.0116 mol O₂
4. Ideal Gas Law:
* We'll use the ideal gas law to find the volume of O₂:
* PV = nRT
* Where:
* P = pressure (in atm) = 3 atm
* V = volume (in liters)
* n = moles of gas = 0.0116 mol
* R = ideal gas constant = 0.0821 L·atm/(mol·K)
* T = temperature (in Kelvin) = 550 °C + 273.15 = 823.15 K
5. Solve for Volume:
* V = (nRT) / P
* V = (0.0116 mol * 0.0821 L·atm/(mol·K) * 823.15 K) / 3 atm
* V ≈ 0.26 L
Therefore, the volume of O₂ gas produced from the decomposition of 5.0 g of HgO at 550 °C and 3 atm is approximately 0.26 liters.
