If 15.22 g of mercury (II) oxide decompose, how many moles of oxygen can be produced?
Have you farted a little too hard and poo poo came out of your booty?
To determine the number of moles of oxygen that can be produced when 15.22 g of mercury (II) oxide decomposes, we need to use the molar mass of mercury (II) oxide.
Step 1: Find the molar mass of mercury (II) oxide.
The molar mass of mercury (II) oxide (HgO) can be found by adding up the atomic masses of the elements in the compound.
Atomic mass of Hg = 200.59 g/mol
Atomic mass of O = 16.00 g/mol
So, the molar mass of HgO is: (200.59 g/mol) + (16.00 g/mol) = 216.59 g/mol.
Step 2: Calculate the number of moles of HgO.
To calculate the number of moles, we use the formula:
moles = mass / molar mass
moles of HgO = 15.22 g / 216.59 g/mol ≈ 0.0704 mol.
Step 3: Determine the moles of O produced.
From the balanced chemical equation for the decomposition of HgO, we know that 2 moles of HgO produce 1 mole of O2.
So, the moles of O2 produced = 0.0704 mol HgO × (1 mol O2 / 2 mol HgO) = 0.0352 mol.
Therefore, approximately 0.0352 moles of oxygen can be produced from the decomposition of 15.22 g of mercury (II) oxide.
To determine the number of moles of oxygen produced when 15.22 g of mercury (II) oxide decomposes, you need to use the molar mass of mercury (II) oxide.
Step 1: Find the molar mass of mercury (II) oxide.
Mercury (II) oxide has the chemical formula HgO, which consists of one atom of mercury (Hg) and one atom of oxygen (O). The molar mass of mercury is 200.59 g/mol, and the molar mass of oxygen is 16.00 g/mol.
To calculate the molar mass of mercury (II) oxide, add the molar masses of mercury and oxygen:
Molar mass of HgO = (1 mol Hg × 200.59 g/mol) + (1 mol O × 16.00 g/mol)
Molar mass of HgO = 200.59 g/mol + 16.00 g/mol
Molar mass of HgO = 216.59 g/mol
Step 2: Convert the given mass of mercury (II) oxide to moles.
To convert grams to moles, divide the given mass by the molar mass:
15.22 g HgO × (1 mol HgO / 216.59 g HgO) = 0.0703 mol HgO
Step 3: Determine the stoichiometry of the reaction.
The balanced chemical equation for the decomposition of mercury (II) oxide is:
HgO → Hg + 1/2 O2
This equation tells us that 1 mole of mercury (II) oxide produces 1/2 mole of oxygen. Therefore, the molar ratio is 1:1/2 or 2:1.
Step 4: Calculate the moles of oxygen produced.
Since the stoichiometry of the reaction tells us that 1 mole of mercury (II) oxide produces 1/2 mole of oxygen, we can multiply the number of moles of mercury (II) oxide by the ratio:
0.0703 mol HgO × (1/2 mol O2 / 1 mol HgO) = 0.0352 mol O2
Therefore, when 15.22 g of mercury (II) oxide decomposes, 0.0352 moles of oxygen can be produced.
2HgO ==> 2Hg + O2
mols HgO = 15.22/molar mass HgO
Using the coefficients in the balanced equation, convert mols Hg to mols O2.