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

1 Answer
anor277
Jan 3, 2018

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

Explanation:

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"#

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