Pertanyaan

Setarakan reaksi berikut dengan metode biloks atau ion elektron! As 2 S 3 + HNO 3 → H 3 AsO 4 + H 2 SO 4 + NO 2 + H 2 O

Setarakan reaksi berikut dengan metode biloks atau ion elektron!

$As_{2} S_{3} + HNO_{3} \to H_{3} AsO_{4} + H_{2} SO_{4} + NO_{2} + H_{2} O$

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Jawaban terverifikasi

Jawaban

reaksi setaranya adalah:

reaksi setaranya adalah: As subscript 2 S subscript 3 and 28 H N O subscript 3 yields 2 H subscript 3 As O subscript 4 and 3 H subscript 2 S O subscript 4 and 28 N O subscript 2 and 8 H subscript 2 O

As subscript 2 S subscript 3 and 28 H N O subscript 3 yields 2 H subscript 3 As O subscript 4 and 3 H subscript 2 S O subscript 4 and 28 N O subscript 2 and 8 H subscript 2 O

Pembahasan

Metode Biloks Menentukan biloks dan menyetarakan jumlah atom: Menyetarakan jumlah hidrogen: Metode Ion Elektron Reduksi: Oksidasi: Mengeliminasi elektron dengan mengkalikan 28 pada reaksi reduksi (karena jumlah elektron pada reaksi oksidasi adalah 28 elektron). Sehingga reaksi ionnya menjadi: Jadi, reaksi setaranya adalah:

Metode Biloks

Menentukan biloks dan menyetarakan jumlah atom:

As subscript 2 S subscript 3 and H N O subscript 3 yields 2 H subscript 3 As O subscript 4 and 3 H subscript 2 S O subscript 4 and N O subscript 2 and H subscript 2 O

Menyetarakan jumlah hidrogen:

As subscript 2 S subscript 3 and 28 H N O subscript 3 yields 2 H subscript 3 As O subscript 4 and 3 H subscript 2 S O subscript 4 and 28 N O subscript 2 and 8 H subscript 2 O

Metode Ion Elektron

As subscript 2 S subscript 3 and H N O subscript 3 yields H subscript 3 As O subscript 4 and H subscript 2 S O subscript 4 and N O subscript 2 and H subscript 2 O

Reduksi:

2 H to the power of plus sign plus N O subscript 3 to the power of minus sign plus e to the power of minus sign yields N O subscript 2 and H subscript 2 O

Oksidasi:

2 As to the power of 3 plus sign and 8 H subscript 2 O yields 2 As O subscript 4 to the power of 3 minus sign end exponent plus 16 H to the power of plus sign and 10 e to the power of minus sign

3 S to the power of 2 minus sign and 12 H subscript 2 O yields 3 S O subscript 4 to the power of 2 minus sign end exponent plus 24 H to the power of plus sign and 18 e to the power of minus sign

Mengeliminasi elektron dengan mengkalikan 28 pada reaksi reduksi (karena jumlah elektron pada reaksi oksidasi adalah 28 elektron). Sehingga reaksi ionnya menjadi:

3 S to the power of 2 minus sign and 2 As to the power of 3 plus sign plus 28 N O subscript 3 to the power of minus sign plus 16 H to the power of plus sign yields 3 S O subscript 4 to the power of 2 minus sign end exponent plus 2 As O subscript 4 to the power of 3 minus sign end exponent plus 28 N O subscript 2 and 8 H subscript 2 O

Jadi, reaksi setaranya adalah: As subscript 2 S subscript 3 and 28 H N O subscript 3 yields 2 H subscript 3 As O subscript 4 and 3 H subscript 2 S O subscript 4 and 28 N O subscript 2 and 8 H subscript 2 O