determine the volume of oxygen calculated at STP which is liberated when 43,4g mercury oxide decomposes completely into mercury and oxygen
2HgO ==> 2Hg + O2
mols HgO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols HgO to mols O2.
Then L O2 = mols O2 x 22.4 L/mol = ?
To determine the volume of oxygen liberated when mercury oxide decomposes completely, we first need to calculate the number of moles of mercury oxide present.
1. Calculate the molar mass of mercury oxide (HgO):
- The molar mass of mercury (Hg) is 200.59 g/mol.
- The molar mass of oxygen (O) is 16.00 g/mol.
- Therefore, the molar mass of mercury oxide (HgO) is 200.59 g/mol + 16.00 g/mol = 216.59 g/mol.
2. Determine the moles of mercury oxide:
- Divide the given mass of mercury oxide (43.4 g) by its molar mass (216.59 g/mol):
- Moles = Mass / Molar mass = 43.4 g / 216.59 g/mol ≈ 0.2001 mol (rounded to four decimal places).
3. Apply the stoichiometry of the balanced chemical equation to determine the moles of oxygen:
The balanced chemical equation for the decomposition of mercury oxide (HgO) is:
2HgO → 2Hg + O2
- According to the balanced equation, 2 moles of mercury oxide produce 1 mole of oxygen. Therefore, the 0.2001 moles of mercury oxide will produce 0.2001 * 1/2 = 0.1000 moles of oxygen.
4. Calculate the volume of oxygen at STP (Standard Temperature and Pressure):
- At STP, the volume of 1 mole of gas is 22.4 liters.
- Therefore, the volume of 0.1000 moles of oxygen will be 0.1000 * 22.4 = 2.24 liters.
Therefore, the volume of oxygen liberated when 43.4 g of mercury oxide decomposes completely at STP is approximately 2.24 liters.