Question #4056c

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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 $(H_{2} O)$ $(H_2O)$ is (1+1) + 16 = 18

Ratio in moles is:

51.35/342 : 48.55/18

$\Rightarrow$ $rArr$ 0.15 : 2.7

$\Rightarrow$ $rArr$ 1 : 18

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

$A l_{2} {(S O_{4})}_{3} .18 H_{2} O$ $Al_2(SO_4)_(3).18H_2O$