What is the percentage by mass of oxygen in Al2(SO4)3. 2H2O?
[Al = 27, S = 32, H = 1, O = 16]
14.29%
25.39%
50.79%
59.25%
Explanation
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Discussions (21)

Al2(SO4)3.2H2O=
27×2+(32+16×4)3+2(2+16)=378
% by masa of O2
16×4×3+16+2=
192+32=224
224/378×100=59.25%

%centage Composition Of Any Element=molar Mass Of The Element/molar Mass Of The Compound * 100. Which Means Molar Mass Of The Compound Al2(SO4)3.2H2O=27*2 + 32*3 + 16*4*3 + 2*1*2 + 2*16=54+96+192+4+32=378g/mol,Therefore Molar Mass Of The Element=192+32=224
That Is, Percentage Composition Of Oxygen In Al2(SO4)3.2H2O=224/378 * 100=22400/378=59.25% The Correct Answer Is (D) Thanks

You are missing out the oxygen in the water molecule which is 32
All the oxygen in the sulphur oxide Is 4×3 ×16 which is 192 adding that of water its total of 224

This question is absolutely wrong....the Answer I mean, what's wrong with this app sef

D is correct. we have to consider Oxygen of the water of crystalization also

Al2(SO4)3. 2H2O?
[Al = 27, S = 32, H = 1, O = 16]
Al=2x27=54
S=3X32=96
O=12X16=192
4x1+2x16=36 (2H2O)
54+96+192+36=378
So Al=54/378x100=14.3%
Sulphur=96/378x100=25.4%
Oxygen=192/378x100= 50.79%
Therefore the percentage by mass of oxygen is 50.79% (C)ans.

How is the mass of o2 224?...wrong solutions... Even your answers doesn't rhyme with your solutions... Correct answer there is b.. Thanks

The total mass of oxygen in the reaction I.e Al2(SO4)3.2H2O is not 224 but is 192 because 4*3 is 12 not 14 therefore the correct option is C




