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A chemist is producing copper(II) chloride using the following reaction:
Cu(s) + Cl2(g) --> CuCl2(s)
In one reaction, the chemist uses 15.0 g of copper metal and 10.0 g of chlorine gas as reactants. When the reaction reaches completion, 16.0 g of copper(II) chloride has been produced. What was the percent yield of the reaction?

  • Chemistry - ,

    must be each reactant ?

    or total reactants ?

    Ill do each

    Step one:

    mm of products
    134. 45g/mol

    step two
    mol of products

    16.0 g/134.45=1.19 x 10^-1 mol x

    step two

    find the lr

    cu=15.0 /63.546 =.236mol produced


    lr is cl


    turn it into grams

    .141mol x 134. 45g/mol=18.96 g

    four % yield

    16.0g/18.96 g x 100=84.4% around.

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