Answer:
This is a ratio of two small integers, 1.500 = 3:2. Let's assume that in the molecule of ferrous oxide there is one oxygen atom and one iron atom, and the ratio of the mass of the oxygen atom to the mass of the iron atom is 0.2865.
Explanation:
Molecular weight of Iron (II) oxide= FeO Molecular weight of Iron (III) oxide= Fe2O3
In FeO , every one iron atom contains one oxygen atom.
In Fe2O3, every one iron atom contains 1.5 oxygen atoms.
So FeO(II) will contain more Iron than FeO(III).
Calculation of percentage of iron is not necessary. I think this argument
alone is enough to prove this.
However on calculating the % of Fe, in FeO and Fe2O3 we get,
% of Iron in FeO= 55.85/(55.85+16) =55.85/71.85= 77.73%
% of Iron in Fe2O3= (2 x55.85)/(111.7+48)
=111.7/159.48 = 69.94%
Answer:
Ok, so the process here is to convert the mass of H2 (hydrogen gas) to moles by dividing the mass by the molar mass of H2. Once you have the moles then you have to multiply by the STP (standard temperature and pressure) molar volume which should be 22.4.
Molar mass of H2 = (1.01)x2 = 2.02g/mol
19.3/2.02 = 9.55 moles
Now just multiply the moles by the molar volume
9.55 moles x 22.4 = 213.92 Litres of H2 are in 19.3g of H2
It could get bigger or it could have something a lot different to it than anything else
Mendeleev organized it by the atomic mass and mosely did it by atomic number