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0.2550 g AlC3 (1 mol/132 g) =0.001932 mol AlCl3

0.001932 mol AlCl3 (6.022 x 10^23 molecules AlCl3/1 mol AlCl3) = 1.163 x 10^21

1.163x10^21 molecules AlCl3 (3 mol Cl/1 mol AlCl3) =3.490x10^21 Cl ions

3.490x10^21 Cl ions (1 mol/6.022 x 10^23) =5.795x10^-3 moles Cl

The formula to solve this problem appears above.

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