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Pada reaksi:  Berapa liter gas   (P,T) yang dihasi...

Pada reaksi: Mg and H subscript 2 S O subscript 4 yields Mg S O subscript 4 and H subscript 2 Berapa liter gas space H subscript 2  (P,T) yang dihasilkan jika 19,2 gram logam magnesium direaksikan dengan larutan asam sulfat bila gas ini diukur pada keadaan dimana 0,5 liter gas O subscript 2  (P,T) = 1,28 gram.

Jawaban:

Mg and H subscript 2 S O subscript 4 yields Mg S O subscript 4 and H subscript 2

Menentukan mol Mg

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Mg end cell equals cell fraction numerator massa space Mg over denominator Ar space Mg end fraction end cell row blank equals cell fraction numerator 19 comma 2 space g over denominator 24 space g forward slash mol end fraction end cell row blank equals cell 0 comma 8 space mol end cell row blank blank blank end table

Menentukan mol gas O subscript 2

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space O subscript 2 end cell equals cell fraction numerator massa space O subscript 2 over denominator Mr space O subscript 2 end fraction end cell row blank equals cell fraction numerator 1 comma 28 space g over denominator 32 space g forward slash mol end fraction end cell row blank equals cell 0 comma 04 space mol end cell end table

Volume gas O subscript 2 = 0,5 L

Menentukan volume Mg

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses n over V close parentheses subscript Mg end cell equals cell open parentheses n over V close parentheses subscript O subscript 2 end subscript end cell row cell fraction numerator 0 comma 8 space mol over denominator V end fraction end cell equals cell fraction numerator 0 comma 04 space mol over denominator 0 comma 5 space L end fraction end cell row V equals cell fraction numerator 0 comma 8 space mol space x space 0 comma 5 space L over denominator 0 comma 04 space mol end fraction end cell row blank equals cell 10 space L end cell row blank blank blank end table

Mg space space space space space plus space space space H subscript 2 S O subscript 4 space space space rightwards arrow space space space Mg S O subscript 4 space space plus space space space H subscript 2 0 comma 2 space mol 

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space H subscript 2 end cell equals cell fraction numerator koefisien space H subscript 2 over denominator koefisien space Mg end fraction x space mol space Mg end cell row blank equals cell 1 over 1 x space 0 comma 2 space mol end cell row blank equals cell 0 comma 2 space mol end cell row cell mol space Mg end cell equals cell mol space H subscript 2 space maka space volume space H subscript 2 equals 10 space L end cell row blank blank blank end table

Jadi, volume gas space H subscript 2 yang dihasilkan adalah 10 L.

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