What is the percentage by

mass of oxygen in
Al2(SO4)3.2H2O?
A. 14.29%
B. 2539%
C. 50.79%
D. 59.25%
(Al = 27, S = 32, H=1, O=16)

3 answers

First, let's calculate the molar mass of Al2(SO4)3.2H2O.

The molar mass of Al2(SO4)3 can be calculated as follows:
2(Al) + 3(S) + 12(O)
= 2(27) + 3(32) + 12(16)
= 54 + 96 + 192
= 342 g/mol

The molar mass of 2H2O can be calculated as follows:
2(2(H) + 16(O))
= 2(2 + 16)
= 2(18)
= 36 g/mol

So, the total molar mass of Al2(SO4)3.2H2O is:
342 + 36
= 378 g/mol

Now, let's calculate the molar mass of oxygen in Al2(SO4)3.2H2O.

The molar mass of oxygen in Al2(SO4)3.2H2O can be calculated as follows:
16(O) in Al2(SO4)3 + 16(O) in 2H2O
= 16 + 16
= 32 g/mol

Finally, let's calculate the percentage by mass of oxygen in Al2(SO4)3.2H2O.

Percentage by mass of oxygen = (molar mass of oxygen / total molar mass of Al2(SO4)3.2H2O) x 100%
= (32 g/mol / 378 g/mol) x 100%
≈ 8.47%

Therefore, the correct answer is not provided.
The Bot is wrong again.
The molar mass of Al2(SO4)3.2H2O is correctly listed as 378.
How many O atoms are in the molecule. There are 3*4 in the (SO4)3 + 2 in 2H2O = 14
The mass of the 14 O atoms = 14*16 = 224
mass % O in Al2(SO4)3.2H2O = (224/378)*100 = 59.25% which is answer D in the problem.
You are correct, and I apologize for the mistake in my previous response. The correct answer is indeed D. The percentage by mass of oxygen in Al2(SO4)3.2H2O is 59.25%. Thank you for pointing out the error.