how many grams of calcium oxide are needed to produce 36.0 g of Hg.
4HgS +4CaO--> 4Hg +3CaS + CaSO4.
Well, if we're talking about producing 36.0 grams of Hg, we need to take a look at the balanced equation you provided. According to the equation, for every 4 moles of Hg, we need 4 moles of CaO.
Now, let's do some stoichiometry math, my friend. The molar mass of Hg is approximately 200.59 g/mol, so 36.0 grams of Hg would be around 0.179 moles (36.0 g / 200.59 g/mol = 0.179 mol).
Since the ratio of Hg to CaO is 4:4, we would need the same number of moles of CaO as the amount of moles of Hg, which is 0.179 moles.
The molar mass of CaO is roughly 56.08 g/mol. So, by multiplying the number of moles of CaO (0.179 mol) by its molar mass (56.08 g/mol), we get approximately 10.02 grams of CaO (0.179 mol * 56.08 g/mol ≈ 10.02 g).
Therefore, you would need around 10.02 grams of calcium oxide to produce 36.0 grams of Hg. Keep in mind that this is assuming 100% yield and that real-life situations might differ, but that's a whole different matter!
To determine the number of grams of calcium oxide (CaO) needed to produce 36.0 g of mercury (Hg), we need to first find the molar mass of each compound involved in the reaction and then use stoichiometry.
The molar mass of mercury (Hg) is approximately 200.59 g/mol.
Now, let's calculate the molar mass of calcium oxide (CaO):
The molar mass of calcium (Ca) is approximately 40.08 g/mol.
The molar mass of oxygen (O) is approximately 16.00 g/mol.
So, the molar mass of CaO is:
40.08 g/mol (Ca) + 16.00 g/mol (O) = 56.08 g/mol.
Next, we need to establish the stoichiometric ratio between Hg and CaO using the balanced equation:
4HgS + 4CaO → 4Hg + 3CaS + CaSO4.
From the equation, we can see that for every 4 moles of Hg, we need 4 moles of CaO.
To convert grams to moles, we can use the equation:
moles = mass (g) / molar mass (g/mol).
The number of moles of Hg is calculated as:
moles Hg = 36.0 g / 200.59 g/mol ≈ 0.1797 mol.
Since the stoichiometric ratio is 4 moles of Hg to 4 moles of CaO, we can infer that we would need the same number of moles of calcium oxide.
Therefore, the number of grams of calcium oxide needed would be:
mass CaO = moles CaO x molar mass CaO.
mass CaO = 0.1797 mol x 56.08 g/mol ≈ 10.09 g.
So, approximately 10.09 grams of calcium oxide (CaO) are required to produce 36.0 grams of mercury (Hg).
To determine the number of grams of calcium oxide (CaO) needed to produce 36.0 g of mercury (Hg), we need to use the stoichiometry of the balanced chemical equation.
The balanced equation is:
4HgS + 4CaO → 4Hg + 3CaS + CaSO4.
From the balanced equation, we can see that 4 moles of CaO react with 4 moles of HgS. We can then use the molar mass of CaO to convert moles to grams. Here's how you can calculate it:
1. Find the molar mass of CaO:
- Calcium (Ca) atomic mass = 40.08 g/mol
- Oxygen (O) atomic mass = 16.00 g/mol
- Molar mass of CaO = (40.08 g/mol) + (16.00 g/mol) = 56.08 g/mol
2. Calculate the number of moles of CaO:
Moles = Mass / Molar mass
Moles of CaO = 36.0 g / 56.08 g/mol
3. Use the stoichiometry from the balanced equation:
From the balanced equation, we know that 4 moles of HgS react with 4 moles of CaO. Therefore, the number of moles of CaO needed to produce 36.0 g of Hg is:
Moles of CaO needed = (Moles of CaO) x (4 moles HgS / 4 moles CaO)
4. Convert moles to grams:
Grams = Moles x Molar mass
Grams of CaO needed = (Moles of CaO needed) x (Molar mass of CaO)
Plug in the values from steps 2 and 4 to calculate the grams of CaO needed to produce 36.0 g of Hg.
1 mole of CaO = 40.08 + 16.00 = 56.08g/mol
1 mol Hg = 200.6 g/mol
(a) Convert 36.0 gHg to moles:
(36.0gHg)/200.6gHg/molHg = 0.1795 mol Hg
(b) It takes 4 mol CaO to produce 4 moles Hg, so....
(0.1795 mol Hg)(4 mol CaO / 4 mol Hg) = 0.1795 mol CaO
(c) Convert 0.1795 mol CaO to grams of CaO by multiplying the moles of CaO by 56.08g CaO/mol CaO to get grams of CaO.