KCL that is contaminated with calcium chlorate (Ca(ClO3)2) is heat in a tube. wat is the % weight of...?

The test tube is heated prompting the reaction:

_Ca(ClO3)2 (s) + _KCl(s) --> _CaCl2(s) + _O2(g) + _KCl(s)

Data during experiment:

Mass of test tube: 7.918g

Mass of mixture (KCl/Ca(ClO3)2) + tube : 19.164g

Mass of residue (KCl/Ca(CaCl2) + tube after heating: 18.635

How do I find the % weight of Ca from (KCl/Ca(ClO3)2)?

and % weight of KCl in original mixture?

Ca: 40.08 g/mol

ClO3: 83.45 g/mol

K: 39.1 g/mol

Cl: 35.45 g/mol

O2: 32.00 g/mol

2 Answers

  • Dr.A
    Lv 7
    1 decade ago
    Favorite Answer

    Ca(ClO3)2 + KCl = CaCl2 + 3 O2 + KCl

    mass mixture KCl / Ca(ClO3)2 = 19.164 - 7.918=11.246 g

    mass KCl/Ca(ClO3)2 after heating = 18.635 - 7.918 =10.717 g

    11.246 - 10.717 =0.529 g = mass O2 lost in the decomposition of Ca(ClO3)2

    moles O2 = 0.529 / 31.998 g/mol=0.0165

    the ratio between Ca(ClO3)2 and O2 is 1 : 3

    moles Ca(ClO2)2 = 0.0165 x 1 /3=0.00550

    = moles Ca

    mass Ca = 0.00550 x 40.08 g/mol=0.220 g

    % Ca = 0.220 x 100/11.246 =1.96

    mass Ca(ClO3)2 = 0.00550 mol x 206.98 g/mol=1.14 g

    mass KCl = 11.246 - 1.14 =10.1 g

    % KCl = 10.1 x 100/ 11.246 =89.9

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