How many grams of fe is present in 30g of FeSO4
30 x (atomic mass Fe/molar mass FeSO4) = ?
how many moles are there in 30g fe SO4
To determine how many grams of Fe (iron) is present in 30g of FeSO4 (iron(II) sulfate), we need to consider the molar mass and stoichiometry of the compound.
The molar mass of FeSO4 can be calculated by adding up the atomic masses of its constituent elements:
Fe: 55.845 g/mol
S: 32.06 g/mol
O: 16.00 g/mol (there are 4 oxygen atoms)
So, the molar mass of FeSO4 is: 55.845 + 32.06 + (4 × 16.00) = 151.91 g/mol
Next, we need to determine the ratio of Fe to FeSO4 using the stoichiometry from the chemical formula. In FeSO4, there is only one Fe atom per FeSO4 molecule.
Now we can set up a proportion to find the mass of Fe in 30g of FeSO4:
(30 g FeSO4) × (1 mol FeSO4/151.91 g FeSO4) × (1 mol Fe/1 mol FeSO4) × (55.845 g Fe/1 mol Fe) = X g Fe
Simplifying the expression gives:
X = (30 × 55.845) / 151.91 ≈ 10.97 g
Therefore, approximately 10.97 grams of Fe is present in 30 grams of FeSO4.