1 Answer

Ceztko · Answer 1 · 2021-08-20T10:24:17+0000

The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)

13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :

$\tt mol=(mass)/(Ar)=\\\\mol=(13.02~g)/(58.933)\\\\mol=0.221$

Mass of metal iodide formed : 97.12 g, so mass of Iodine :

$\tt =mass~metal~iodide-mass~Cobalt\\\\=97.12-13.02\\\\=84.1~g$

Then mol iodine (MW=126.9045 g/mol) :

$\tt (84.1)/(126.9045)=0.663$

mol ratio of Cobalt and Iodine in the compound :

$\tt 0.221/ 0.663=1/ 3$

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