In 1774 Joseph Priestly conducted one of his most famous experiments which
lead to a method for the preparation of oxygen. The experiment involved heating
a sample of mercury II oxide with a large lens.
The equation for this reaction is shown below:
2HgO(s) = 2Hg(l) + O2(g)
a) What volume of O2(g) would be obtained if 1.08g of mercury
II oxide were completely decomposed? (Given that 1 mole of a gas occupies 24
dm3 under the experimental conditions)
Here is a worked example. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
According to the laws of thermodynamics, if you heat a large enought object, it gets enough energy it implies. What whould happen if you add 80 joules of energy to 1kg of zinc metal.... i just dont get it
To calculate the volume of oxygen gas (O2) produced in this reaction, we need to use the given information and the stoichiometry of the reaction.
1. Determine the molar mass of mercury II oxide (HgO):
- Mercury (Hg) has a molar mass of 200.59 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- Since there are two oxygen atoms in each molecule of HgO, the molar mass of HgO is: 2 * 16.00 g/mol + 200.59 g/mol = 232.59 g/mol.
2. Calculate the number of moles of HgO in 1.08 g:
- Number of moles = mass / molar mass = 1.08 g / 232.59 g/mol.
3. Determine the stoichiometric ratio between HgO and O2 in the balanced equation:
- According to the balanced equation, 2 moles of HgO produce 1 mole of O2.
4. Calculate the number of moles of O2 produced from the given mass of HgO:
- Number of moles of O2 = number of moles of HgO * (1 mole of O2 / 2 moles of HgO).
5. Convert the number of moles of O2 to volume using the given information:
- Given that 1 mole of a gas occupies 24 dm3, the volume of O2 gas produced would be: number of moles of O2 * 24 dm3.
Now, let's calculate step by step:
1. Molar mass of HgO:
- Molar mass of HgO = 2 * 16.00 g/mol + 200.59 g/mol = 232.59 g/mol.
2. Number of moles of HgO:
- Number of moles of HgO = 1.08 g / 232.59 g/mol.
3. Stoichiometric ratio:
- 2 moles of HgO produce 1 mole of O2.
4. Number of moles of O2:
- Number of moles of O2 = (number of moles of HgO) * (1 mole of O2 / 2 moles of HgO).
5. Volume of O2:
- Volume of O2 = (number of moles of O2) * 24 dm3.
After calculating these steps, you will get the volume of O2 gas produced from 1.08 g of mercury II oxide.