How many moles of ammonium sulfate were released in 25.3 kg?
25.3 kg of what? (NH4)2SO4?
mols = grams/molar mass.
To determine the number of moles of ammonium sulfate released in 25.3 kg, we need to use the molar mass of ammonium sulfate and the given mass.
The molar mass of ammonium sulfate (NH4)2SO4 can be calculated by summing the atomic masses of its constituent elements:
N = 14.01 g/mol (Nitrogen)
H = 1.008 g/mol (Hydrogen)
S = 32.07 g/mol (Sulfur)
O = 16.00 g/mol (Oxygen)
Molar mass of (NH4)2SO4 = 14.01 * 2 + 1.008 * 8 + 32.07 + 16.00 * 4 = 132.14 g/mol
Now, to find the number of moles, we can use the equation:
moles = mass / molar mass
Substituting the known values:
moles = 25.3 kg / (132.14 g/mol)
But since we have the mass in kg and the molar mass in g/mol, we need to convert the mass to grams:
25.3 kg = 25.3 * 1000 g = 25300 g
Now, calculating the moles:
moles = 25300 g / 132.14 g/mol
moles ≈ 191.51 mol
Therefore, approximately 191.51 moles of ammonium sulfate were released in 25.3 kg.