If You Heated 100 Grams Of Powdered Mercuric Oxide To Produce 93 Grams Of Liquid Mercury How Much Oxygen Would Be Released
2HgO ==> 2Hg + O2
mols HgO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols HgO to mols O2.
Now convert mols O2 to whatever unit you want. The post doesn't say. If grams, then grams O2 = mols O2 x molar mass O2.
If volume at STP, then L = mols x 22.4 L/mol = ?
I really don't know im not smart
heated 100 grams of powdered mercuric oxide to produce 93 grams of liquid mercury, how much oxygen would be released?
To determine the amount of oxygen that would be released when heating powdered mercuric oxide to produce liquid mercury, we first need to calculate the difference in mass between the two substances.
The chemical equation for the reaction is as follows:
2 HgO(s) -> 2 Hg(l) + O2(g)
From the equation, we can see that for every 2 moles of mercuric oxide (HgO), 1 mole of oxygen gas (O2) is produced. We need to find the moles of mercuric oxide used, and then use the stoichiometry of the reaction to determine the moles of oxygen gas:
1. Calculate the moles of mercuric oxide (HgO):
Moles of HgO = Mass of HgO / Molar mass of HgO
The molar mass of HgO can be calculated by summing the atomic masses of its constituents, which are mercury (Hg) and oxygen (O). According to the periodic table:
Atomic mass of Hg = 200.59 g/mol
Atomic mass of O = 16.00 g/mol
Molar mass of HgO = Atomic mass of Hg + Atomic mass of O = 200.59 + 16.00 = 216.59 g/mol
Therefore, the moles of HgO = 100 g / 216.59 g/mol.
2. Use the stoichiometry of the reaction to determine the moles of oxygen gas produced:
Moles of O2 = Moles of HgO (according to the balanced equation)
Finally, to find the mass of oxygen gas released, we can use the equation:
Mass of O2 = Moles of O2 x Molar mass of O2
The molar mass of O2 is approximately 32.00 g/mol.
By plugging in the values obtained from the calculations above into the equation, we can determine the mass of oxygen gas released during the reaction.