Roboguru

setarakan reaksi redoks berikut. (suasana basa)

setarakan reaksi redoks berikut.


begin mathsize 14px style Zn open parentheses italic s close parentheses and N O subscript 3 to the power of minus sign open parentheses italic a italic q close parentheses yields Zn O subscript 2 to the power of 2 minus sign open parentheses italic a italic q close parentheses plus N H subscript 3 open parentheses italic g close parentheses end style (suasana basa)

  1. ... space  

  2. ... undefined 

Jawaban:

Tentukan atom yang mengalami perubahan bilangan oksidasi dan setarakan jumlah atom yang berubah bilangan oksidasinya.undefined 

undefined 

Kalikan pengali dengan koefisien:undefined 

begin mathsize 14px style bold 4 Zn open parentheses italic s close parentheses plus N O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis rightwards arrow bold 4 Zn O subscript 2 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis plus N H subscript 3 open parentheses italic g close parentheses end style 

Setarakan jumlah O dengan menambahkan undefined:undefined 

begin mathsize 14px style 4 Zn open parentheses italic s close parentheses plus N O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis plus bold 5 H subscript bold 2 O bold open parentheses italic l bold close parentheses yields 4 Zn O subscript 2 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis plus N H subscript 3 open parentheses italic g close parentheses end style 

Setarakan jumlah H dengan menambahkan ion undefined:undefined 

begin mathsize 14px style 4 Zn open parentheses italic s close parentheses plus N O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis plus 5 H subscript 2 O open parentheses italic l close parentheses yields 4 Zn O subscript 2 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis plus N H subscript 3 open parentheses italic g close parentheses and bold 7 H to the power of bold plus sign bold left parenthesis italic a italic q bold right parenthesis end style 

Tambahkan ion undefined di kedua ruas sebanyak jumlah ion undefined:

begin mathsize 14px style 4 Zn open parentheses italic s close parentheses plus N O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis plus 5 H subscript 2 O open parentheses italic l close parentheses and bold 7 O H to the power of bold minus sign bold left parenthesis italic a italic q bold right parenthesis rightwards arrow 4 Zn O subscript 2 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis plus N H subscript 3 open parentheses italic g close parentheses and 7 H to the power of plus sign left parenthesis italic a italic q right parenthesis plus bold 7 O H to the power of bold minus sign bold left parenthesis italic a italic q bold right parenthesis end style 

Ion undefined dan undefined akan membentuk undefined. Eliminasi undefined pada kedua ruas sehingga diperoleh reaksi setara:

begin mathsize 14px style 4 Zn open parentheses italic s close parentheses plus N O subscript 3 to the power of minus sign left parenthesis italic a italic q italic right parenthesis plus 7 O H to the power of minus sign left parenthesis italic a italic q right parenthesis rightwards arrow 4 Zn O subscript 2 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis plus N H subscript 3 open parentheses italic g close parentheses and 2 H subscript 2 O open parentheses italic l close parentheses end style 

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