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18 August, 11:56

How many grams of copper (II) chloride dihydrate would be requires to react completely with 1.20 g of aluminum

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  1. 18 August, 12:09
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    3CuCl2 • 2H2O + Al - - - > 3Cu + 2AlCl3 + 6H2O

    moles of Al = 1.2 / 27 = 0.044

    1 mole of Al requires 3 moles of CuCl2 • 2H2O

    so, 0.044 mole of Al requires = 0.044 x 3 moles of CuCl2 • 2H2O

    = 0.133 moles

    so mass = 0.133 x 170 = 22.61 g
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