Roboguru

Perhatikan reaksi berikut! Menurut Anda, apakah p...

Perhatikan reaksi berikut!

Menurut Anda, apakah persamaan reaksi tersebut sudah setara?  Lakukan pengecekan ulang, jika belum setara, tuliskan persamaan reaksi yang lebih tepat!

begin mathsize 14px style K subscript 4 Fe open parentheses C N close parentheses subscript 6 left parenthesis italic a italic q right parenthesis plus H subscript 2 S O subscript 4 left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses yields K subscript 2 S O subscript 4 left parenthesis italic a italic q right parenthesis plus Fe S O subscript 4 left parenthesis italic a italic q right parenthesis plus open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript left parenthesis italic a italic q right parenthesis plus C O open parentheses italic g close parentheses end style undefined 

Jawaban:

Reaksi tersebut belum setara, bila disetarakan akan seperti berikut:
Reaksi setara apabila jumlah atom produk = atom reaktan
penyetaraan reaksi mengunakan metode biloks dan setengah reaksi.
langkah pertama. tentukan ionisasi dari K subscript 4 Fe open parentheses C N close parentheses subscript 6 comma space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript
K subscript 4 Fe open parentheses C N close parentheses subscript 6 space rightwards arrow space 4 K to the power of plus sign space plus space Fe to the power of 2 plus sign space plus space 6 C N to the power of minus sign open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 space rightwards arrow space 2 N H subscript 4 to the power of plus space plus space S O subscript 4 to the power of 2 minus sign end exponent

Langkah kedua. tentukan yang mengalami redoks
6 C N to the power of minus sign space rightwards arrow space 2 N H subscript 4 to the power of plus space plus space C O

Langkah ketiga. setarakan atom
6 C N to the power of minus sign space rightwards arrow space 6 N H subscript 4 to the power of plus space plus space 6 C O

Langkah keempat. setarakan suasana asam
12 H to the power of plus sign space plus space 6 H subscript 2 O space plus space 6 C N to the power of minus sign space rightwards arrow space 6 N H subscript 4 to the power of plus space plus space 6 C O

Langkah kelima. setarakan ion K to the power of plus sign space Fe to the power of 2 plus sign  dan S O subscript 4 to the power of 2 minus sign end exponent
6 S O subscript 4 to the power of 2 minus sign end exponent space plus space 4 K to the power of plus sign space plus space Fe to the power of 2 plus sign space plus space 12 H to the power of plus sign space plus space 6 H subscript 2 O space plus space 6 C N to the power of minus sign space rightwards arrow space 6 N H subscript 4 to the power of plus space plus space 6 C O space plus space 4 K to the power of plus sign space plus space Fe to the power of 2 plus sign space plus space 6 S O subscript 4 to the power of 2 minus sign end exponent

Langkah keenam. setarakan 

 begin mathsize 14px style K subscript 4 Fe open parentheses C N close parentheses subscript 6 plus 6 H subscript 2 S O subscript 4 and 6 H subscript 2 O yields 6 C O and Fe S O subscript 4 and 3 open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript and 2 K subscript 2 S O subscript 4 end style 

Jadi, jawaban yang benar adalah

 begin mathsize 14px style K subscript 4 Fe open parentheses C N close parentheses subscript 6 plus 6 H subscript 2 S O subscript 4 and 6 H subscript 2 O yields 6 C O and Fe S O subscript 4 and 3 open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript and 2 K subscript 2 S O subscript 4 end style

1

Ruangguru

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved