Do i need a chemical formula to calculate the mols of FeSO4*7H20 when given the mols of Fe(II)?
If so i only know that Fe(III) + H2SO4 + H3PO4 are the reactants, and FeSO4*H2O is the product. Help?
Ok, your first sentence is given the mols of Fe(II). Did you mean Fe(III) ? I assume so.
Balance the equation.
Fe + H2SO4+H3PO4>>FeSO4.7H20 + where did the phosphate go? Your reaction is incomplete
the mols for Fe(II) are the same as Fe(III) since the ratio is 1:1.
And that's the problem with the reaction. I don't know what else is made. I was just given those components. I'm assuming it would yeild FsSO4*7H2O + P? But i am unsure.
The only hint I was given was that I use the mols of Fe(II) to find the mols of FeSO4*7H20
Yes, you do need a chemical formula to calculate the moles of FeSO4·7H2O when given the moles of Fe(II). In this case, you need to determine the stoichiometry of the reaction between Fe(II) and H2SO4 to find the molar ratio between FeSO4·7H2O and Fe(II).
Given that Fe(III) + H2SO4 + H3PO4 are the reactants and FeSO4·H2O is the product, it suggests that Fe(II) is being oxidized to Fe(III) by H2SO4 and H3PO4. However, without further information about the reaction conditions and the overall balanced equation, it is not possible to directly determine the molar ratio between FeSO4·7H2O and Fe(II).
To calculate the moles of FeSO4·7H2O, you would need to know the balanced equation for the reaction and the molar stoichiometry between FeSO4·7H2O and Fe(II) in that equation. Once you have that information, you can use the following steps:
1. Write the balanced equation for the reaction between Fe(II) and the other reactants (H2SO4 and H3PO4).
2. Determine the molar stoichiometry between FeSO4·7H2O and Fe(II) in the balanced equation.
3. Use the given moles of Fe(II) to calculate the moles of FeSO4·7H2O using the molar stoichiometry obtained in step 2.
Note that without specific information about the balanced equation and the molar stoichiometry between FeSO4·7H2O and Fe(II), it is not possible to provide a specific calculation.