I‘m rather stuck on this question:What mass of iron(II) sulfate would be needed to provide 28 grams of iron?I know that the mass of one mole of iron(II) sulfate is 152g, but I‘m not entirely certain of the question.All posts are appreciated!
that is FeSO4. The Roman numeral II shows that the iron is in the +2 oxidation state. as a results of fact the sulfate anion is in the -2 oxidation state, there are one among each of them to balane the costs (internet value 0). Now take the atomic mass of each element and multiply it via the style of atoms of that element and upload all of them as much as get the formula mass for FeSO4. Then multiply that via 0.354 g/mole and you have have been given your answer: (sixty 5.80 5 x a million) + (32.06 x a million) + sixteen.00 x 4) 161.80 one g/mol 161.80 one g/mol x 0.354 mol fifty seven.3 g FeSO4
Iron (II) Sulfate formula is FeSO4 That is 1--Iron,1-Sulfur,4-Oxygen Atomic weights are Fe-56 S-32 O-16 (4 O- 64) Add them up and you get 152 thats why one mole weighs 152 grams So one mole of Iron(II) Sulfate has 56 grams of Iron in it. To get 28 grams of Iron would be one half mole or 76 grams of Iron (II) Sulfate. Make Sense ?
