Question #4056c

Question

1581 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-10-19T23:34:51

The value of n = 18.

The percentage of anhydrous aluminium sulfate must be 100 - 48.65 = 51.35 %

The Mr of aluminium sulfate is (2 x27) +[32 +(4 x16)] x 3 = 342

The Mr of water $(H_2O)$ is (1+1) + 16 = 18

Ratio in moles is:

51.35/342 : 48.55/18

$rArr$ 0.15 : 2.7

$rArr$ 1 : 18

So the formula of hydrated aluminium sulfate , in this case, is

$Al_2(SO_4)_(3).18H_2O$