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DDT dibuat dari reaksi kloral ( CCl 3 ​ − CHO ) dengan klorobenzena ( C 6 ​ H 5 ​ Cl ) . CCl 3 ​ − CHO + 2 C 6 ​ H 5 ​ Cl → CCl 3 ​ − CH ( C 6 ​ H 4 ​ Cl ) 2 ​ + H 2 ​ O Jika 1 kg kloral bereaksi tuntas dengan 1 kg klorobenzena, berapa kg DDT dapat dihasilkan? ( A r ​ H = 1 , C = 12 , O = 16 , dan + 4 m u Cl = 35 , 5 )

DDT dibuat dari reaksi kloral  dengan klorobenzena .

 

Jika 1 kg kloral bereaksi tuntas dengan 1 kg klorobenzena, berapa kg DDT dapat dihasilkan?

 space space space

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DDT yang dihasilkan adalah 1,58 kg.

DDT yang dihasilkan adalah 1,58 kg.space space space

Pembahasan

Penentuan pereaksi pembatas: Karena , maka pereaksi pembatasnya adalah klorobenzena . Reaksi MRS Penentuan massa DDT yang dihasilkan: Jadi, DDT yang dihasilkan adalah 1,58 kg.

  1. Penentuan pereaksi pembatas:
     

    n space kloral equals fraction numerator m space kloral over denominator italic M subscript r space kloral end fraction n space kloral equals fraction numerator 1000 space g over denominator 147 comma 5 space g space mol to the power of negative sign 1 end exponent end fraction n space kloral equals 6 comma 8 space mol  fraction numerator n space kloral over denominator koef point space kloral end fraction equals fraction numerator 6 comma 8 space mol over denominator 1 end fraction fraction numerator n space kloral over denominator koef point space kloral end fraction equals 6 comma 8 space mol   n space C subscript 6 H subscript 5 Cl equals fraction numerator m space C subscript 6 H subscript 5 Cl over denominator italic M subscript r space C subscript 6 H subscript 5 Cl end fraction n space C subscript 6 H subscript 5 Cl equals fraction numerator 1000 space g over denominator 112 comma 5 space g space mol to the power of negative sign 1 end exponent end fraction n space C subscript 6 H subscript 5 Cl equals 8 comma 9 space mol  fraction numerator n space C subscript 6 H subscript 5 Cl over denominator koef point space C subscript 6 H subscript 5 Cl end fraction equals fraction numerator 8 comma 9 space mol over denominator 2 end fraction fraction numerator n space C subscript 6 H subscript 5 Cl over denominator koef point space C subscript 6 H subscript 5 Cl end fraction equals 4 comma 45 space mol   


    Karena fraction numerator n space C subscript 6 H subscript 5 Cl over denominator koef point space C subscript 6 H subscript 5 Cl end fraction less than fraction numerator n space kloral over denominator koef point space kloral end fraction, maka pereaksi pembatasnya adalah klorobenzena open parentheses C subscript 6 H subscript 5 Cl close parentheses.

  2. Reaksi MRS

    space space space space space space C Cl subscript 3 bond C H O space plus space 2 C subscript 6 H subscript 5 Cl space rightwards arrow space C Cl subscript 3 bond C H open parentheses C subscript 6 H subscript 4 Cl close parentheses subscript 2 space plus space space space H subscript 2 O M space space space space space space space 6 comma 8 space mol space space space space space space space space space space 8 comma 9 space mol space space space space space space space space space space space space space space space space space space 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 minus sign bottom enclose R space minus sign begin inline style 1 half end style left parenthesis 8 comma 9 space mol right parenthesis space space space space minus sign 8 comma 9 space mol space space space space space space space plus begin inline style 1 half end style left parenthesis 8 comma 9 space mol right parenthesis space space space space space space space space space space plus begin inline style 1 half end style left parenthesis 8 comma 9 space mol right parenthesis end enclose S space space space space space space space 2 comma 35 space mol space space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space space 4 comma 45 space mol space space space space space space space space space space space space space space space space space 4 comma 45 space mol 

     
  3. Penentuan massa DDT yang dihasilkan:

    m space DDT double bond n space DDT cross times italic M subscript r space DDT m space DDT equals 4 comma 45 space mol cross times 354 comma 5 space g space mol to the power of negative sign 1 end exponent m space DDT equals 1 comma 58 space kg  


Jadi, DDT yang dihasilkan adalah 1,58 kg.space space space

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