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Setarakan reaksi berikut dengan cara setengah reaksi!   K2​Cr2​O7​+SnCl2​+HCl→CrCl3​+SnCl4​+KCl+H2​O

Pertanyaan

Setarakan reaksi berikut dengan cara setengah reaksi!space
 

K subscript 2 Cr subscript 2 O subscript 7 and Sn Cl subscript 2 and H Cl yields Cr Cl subscript 3 and Sn Cl subscript 4 and K Cl and H subscript 2 O

N. Puspita

Master Teacher

Jawaban terverifikasi

Pembahasan

Langkah Pertama membagi reaksi menjadi setengah reaksi reduksi dan oksidasi, cukup dibuatkan reaksi ionnya saja. 

Red space colon space Cr subscript 2 O subscript 7 to the power of minus sign yields 2 Cr to the power of 3 plus sign Oks space colon space Sn to the power of 2 plus sign yields Sn to the power of 4 plus sign

Tambahkan H2O pada ruas yang kekurangan atom O

Red space colon space Cr subscript 2 O subscript 7 to the power of minus sign yields 2 Cr to the power of 3 plus sign and 7 H subscript 2 O Oks space colon space Sn to the power of 2 plus sign yields Sn to the power of 4 plus sign

Tambahkan H+​ pada ruas yang kekurangan atom H

Red space colon space Cr subscript 2 O subscript 7 to the power of minus sign plus 14 H to the power of plus sign yields 2 Cr to the power of 3 plus sign and 7 H subscript 2 O Oks space colon space Sn to the power of 2 plus sign yields Sn to the power of 4 plus sign

Samakan jumlah muatan dengan menambahkan ion elektron

Red space colon space Cr subscript 2 O subscript 7 to the power of minus sign plus 14 H to the power of plus sign and 6 e to the power of minus sign yields 2 Cr to the power of 3 plus sign and 7 H subscript 2 O Oks space colon space Sn to the power of 2 plus sign yields Sn to the power of 4 plus sign and 2 e to the power of minus sign

Berikutnya menyamakan elektronnya dengan mengkali silang jumlah elektronnya:
Red space colon space Cr subscript 2 O subscript 7 to the power of minus sign plus 14 H to the power of plus sign and 6 e to the power of minus sign yields 2 Cr to the power of 3 plus sign and 7 H subscript 2 O space space space space space cross times 1 Oks space colon space Sn to the power of 2 plus sign yields Sn to the power of 4 plus sign and 2 e to the power of minus sign space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space cross times 3

Selanjutnya menjumlahkan reaksi reduksi dan oksidasinya sebagai berikut:
Red space colon space Cr subscript 2 O subscript 7 to the power of minus sign plus 14 H to the power of plus sign and 6 e to the power of minus sign yields 2 Cr to the power of 3 plus sign and 7 H subscript 2 O space space space space space Oks space colon space 3 Sn to the power of 2 plus sign yields 3 Sn to the power of 4 plus sign and 6 e to the power of minus sign space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space

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space space space space space space space space space Cr subscript 2 O subscript 7 to the power of minus sign plus 3 Sn to the power of 2 plus sign and 14 H to the power of plus sign yields 2 Cr to the power of 3 plus sign and 3 Sn to the power of 4 plus sign and 7 H subscript 2 O space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space

Dan terakhir mengembalikan reaksi nya menjadi bentuk semula, maka akan diperoleh reaksinya yang setara yaitu:
 

K subscript 2 Cr subscript 2 O subscript 7 plus 3 Sn Cl subscript 2 and 14 H Cl yields 2 Cr Cl subscript 3 and 3 Sn Cl subscript 4 and 2 K Cl and 7 H subscript 2 O

 

Jadi reaksi setaranya adalah :

K subscript 2 Cr subscript 2 O subscript 7 plus 3 Sn Cl subscript 2 and 14 H Cl yields 2 Cr Cl subscript 3 and 3 Sn Cl subscript 4 and 2 K Cl and 7 H subscript 2 O

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