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3.2g of a mixture of kn03 and nan03 was heated to constant weight which was found to be 2.64g. What is the % kn03 in mixture

the answer 44.22%

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  1. If I get time I will look at this, but really you need a chemistry teacher.

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  2. 2KNO3 ==> 2KNO2 + O2 and
    2NaNO3 ==> 2NaNO2 + O2

    Let X = mass KNO3
    and Y = mass NaNO3
    eqn 1 is X + Y = 3.2
    eqn 2 is
    X(molar mass KNO2/KNO3) + Y(molar mass NaNO2/NaNO3) = 2.64

    Solve the two equations simultaneously to find X, then
    %X = (mass X/3.2)*100 = ?

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