Consider the salt FeSO4.7H20.

What quantity in grams of salt could be produced from 1g of iron, assuming the overall reaction yield is 75% ? :)

Thanks in advance!

To determine the quantity of FeSO4·7H2O that can be produced from 1g of iron, we first need to find the molar mass of FeSO4·7H2O.

The molar mass of FeSO4 is calculated as follows:
Molar mass of FeSO4 = Atomic mass of Fe + Atomic mass of S + 4*(Atomic mass of O)
= 55.845 g/mol + 32.06 g/mol + 4*(16.00 g/mol)
= 151.91 g/mol

The molar mass of H2O is calculated as follows:
Molar mass of H2O = 2*(Atomic mass of H) + Atomic mass of O
= 2*(1.008 g/mol) + 16.00 g/mol
= 18.02 g/mol

Now, we can find the molar mass of FeSO4·7H2O as follows:
Molar mass of FeSO4·7H2O = Molar mass of FeSO4 + 7*(Molar mass of H2O)
= 151.91 g/mol + 7*(18.02 g/mol)
= 277.93 g/mol

Now, let's calculate the moles of FeSO4·7H2O that can be produced from 1g of iron.

Moles of FeSO4·7H2O = Mass of iron (in grams) / Molar mass of iron
= 1g / 55.845 g/mol
= 0.0179 mol

Since the overall reaction yield is given as 75%, we multiply the moles of FeSO4·7H2O by the yield to find the actual moles of FeSO4·7H2O that will be produced:

Actual moles of FeSO4·7H2O = Moles of FeSO4·7H2O * Reaction yield
= 0.0179 mol * 0.75
= 0.013425 mol

Finally, we can calculate the mass of FeSO4·7H2O produced by multiplying the actual moles of FeSO4·7H2O by its molar mass:

Mass of FeSO4·7H2O produced = Actual moles of FeSO4·7H2O * Molar mass of FeSO4·7H2O
= 0.013425 mol * 277.93 g/mol
= 3.721 g

Therefore, approximately 3.721 grams of FeSO4·7H2O can be produced from 1 gram of iron, assuming the overall reaction yield is 75%.