what mass of Silver oxide would be required to produce 100.0 g. of silver.
Ag2O ==> 2Ag + ?
The equation isn't balanced but it lets you know that you can produce 2 mols Ag for each mols Ag2O.
mols Ag2O = grams/molar mass = ?
mols Ag = 2*mols Ag2O
mols Ag = ?
Then g Ag = mols Ag x atomic mass Ag.
To find the mass of Silver oxide required to produce 100.0 g of silver, we need to determine the molar ratio between Silver oxide (Ag2O) and silver (Ag).
The balanced chemical equation for the reaction is:
2 Ag2O -> 4 Ag + O2
From the equation, we can see that two moles of Silver oxide are required to produce four moles of silver.
Now, we need to calculate the molar mass of silver (Ag) and Silver oxide (Ag2O) to use it in our calculations.
The atomic mass of silver (Ag) is approximately 107.87 g/mol.
The atomic mass of oxygen (O) is approximately 15.999 g/mol.
The molar mass of Silver oxide (Ag2O) can be calculated as follows:
2 × (atomic mass of Ag) + atomic mass of O
= 2 × 107.87 g/mol + 15.999 g/mol
≈ 231.739 g/mol
Next, we can use the molar mass and molar ratio to determine the mass of Silver oxide.
1. Determine the moles of silver (Ag) produced:
Given mass of silver (Ag) = 100.0 g
Moles of Ag = mass / molar mass = 100.0 g / 107.87 g/mol ≈ 0.926 mol
2. Use the molar ratio to determine the moles of Silver oxide (Ag2O) needed:
Moles of Ag2O = 2 × Moles of Ag = 2 × 0.926 mol ≈ 1.852 mol
3. Calculate the mass of Silver oxide (Ag2O):
Mass of Ag2O = Moles of Ag2O × Molar mass of Ag2O
= 1.852 mol × 231.739 g/mol ≈ 428.55 g
Therefore, approximately 428.55 grams of Silver oxide (Ag2O) would be required to produce 100.0 grams of silver (Ag).