Question

In 40g de clorat de potasiu impur sunt 5,4198 ×10²³ atomi de oxigen. Calculeaza masa de potasiu si puritatea probei.

Answer 1

M KClO3 = 39+35,5+3*16 = 122,5

122,5 g KClO3......... 3*6,023*10^23 atomi de oxigen
40 g KClO3...............x
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x = 5,9*10^23 atomi de oxigen

5,9*10^23 atomi de oxigen........... 100%
5,4198*10^23 atomi de oxigen.......y
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y = 91,86% e puritatea probei

122,5 g KClO3.............. 39 g K
40  g KClO3....................z
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z = 12,73 g K