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a fost răspuns

40 g clorat de potasiu impur contine 5,4207 ·10²³ atomi de oxigen . Stiind ca impuritatile nu contin oxigen , sa se calculeze cantitatea de clorat de potasiu pur si puritatea probei !! 

Răspuns :

Wizard48
KClO₃=clorat de potasiu

1mol KClO₃.......................3*6.023*10²³ atomi oxigen
x moli.................................5.4207*10²³ atomi oxigen
x=5.4207*10²³*1/3*6.023*10²³=0.3moli KClO₃

Masa molara KClO₃=39+35.5+3*16=122.5g/mol

1mol.....................................122.5g KClO₃
0.3moli....................................y
y=0.3*122.5/1=36.75(cantitatea pura)

puritatea=36.75*100/40
puritatea=91.875%

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