When 1.335g of a chloride of aluminium is added to excess silver nitrate solution 4.305g of silver chloride is?

(continued) Produced. Calculate the empirical formula of the chloride of aluminium.

Please explain workings!

Thanks!!!! :D

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  • 1 decade ago
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    use molar masses to find mass of Cl

    4.305g of AgCl @ (35.45 g/mol Cl) / (143.3 g/mol AgCl) = 1.065 g Cl

    1.335g of a chloride of aluminium - 1.065 g Cl = 0.270 g Al

    kuse molar masses tofind moles:

    0.270 g Al @ 27 g/mol Al = 0.010 mol Al

    1.065 g Cl @ 3..45 g/mol Cl = 0.030 mol Cl

    your answer is

    AlCl3

  • 1 decade ago

    if we're allowed to assume formula for silver choride=AgCl..then 4.305g AgCl=0.03moles (nearly)..in chem rxn eqns whole numbers of moles interact..or simple ratios thereof..this implies 3moles of AgCl..ie 3 Ag atoms requd to balance equn..if we also assume silver nitrate=AgNO3

    ..then X mols of Aluminium Chloride react with 3 mols AgCl..we have 3 nitrate (NO3) species to allocate ..and formula for Al nitrate=Al(NO3)3

    ..therefore..(I can now see why I gave up Chem in 5th yr!)..balanced equn is

    AlCl3+3AgNO3>3AgCl+Al(NO3)3..emp form= AlCl3

    Source(s): madeitup using ratio/proportion and looking up atomic masses like a bl**dy eejit!
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