How many $Cl^(-)$ ions are there in 584.6 grams of $AlCl_3$ ?

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Answer 1 · 2018-01-03T10:09:10

Well, formally there are 3 moles of chloride per mole of aluminum chloride..

And so we (i) assess the molar quantity of aluminum chloride, $=(584.6*g)/(133.34*g*mol^-1)=4.38*mol$ ...

And (ii) we multiply this molar quantity by factors of THREE and Avogadro's number to get the number of chloride ligands...

$=13.15*molxx"avocado number"$

$13.15*molxx6.022xx10^23*mol^-1=7.92xx10^24*"chlorides"$

How many Cl^(-)Cl^(-) ions are there in 584.6 grams of AlCl_3AlCl_3?