Roboguru

Diketahui reaksi:   Jika 10 gram batu kapur direaksikan dengan asam klorida encer, maka pada keadaan di mana 7 gram gas bervolume 10 liter, volume  yang dihasilkan adalah ... ( Ca = 40, O = 16, C = 12)

Pertanyaan

Diketahui reaksi:

begin mathsize 14px style Ca C O subscript 3 open parentheses italic s close parentheses space plus space 2 H C I left parenthesis italic a italic q right parenthesis space rightwards arrow space Ca Cl subscript 2 left parenthesis italic a italic q right parenthesis space plus space H subscript 2 O open parentheses italic l close parentheses space plus space C O subscript 2 open parentheses italic g close parentheses end style 

Jika 10 gram batu kapur direaksikan dengan asam klorida encer, maka pada keadaan di mana 7 gram gas undefinedbervolume 10 liter, volume begin mathsize 14px style C O subscript 2 end style yang dihasilkan adalah ... (begin mathsize 14px style italic A subscript r end style Ca = 40, O = 16, C = 12)

  1. 2 liter space 

  2. 4 liter undefined 

  3. 6 liter undefined 

  4. 8 liter undefined 

  5. 10 liter undefined 

Pembahasan Soal:

Menentukan nilai mol begin mathsize 14px style C O subscript 2 end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space Ca C O subscript 3 space end cell equals cell space left parenthesis 1 cross times italic A subscript r space Ca right parenthesis plus left parenthesis 1 cross times italic A subscript r space C right parenthesis plus left parenthesis 3 cross times italic A subscript r space O right parenthesis end cell row blank equals cell space left parenthesis 1 cross times 40 right parenthesis plus left parenthesis 1 cross times 12 right parenthesis plus left parenthesis 3 cross times 16 right parenthesis end cell row blank equals cell space 40 plus 12 plus 48 end cell row blank equals cell space 100 space g space mol to the power of negative sign 1 end exponent end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript Ca C O subscript 3 end subscript space end cell equals cell space m over italic M subscript r end cell row blank equals cell space 10 over 100 end cell row blank equals cell space 0 comma 1 space mol end cell end table end style  

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript C O subscript 2 end subscript space end cell equals cell space fraction numerator Koef space C O subscript 2 over denominator Koef space Ca C O subscript 3 end fraction cross times n subscript Ca C O subscript 3 end subscript end cell row blank equals cell space 1 over 1 cross times 0 comma 1 end cell row blank equals cell space 0 comma 1 space mol end cell end table end style 

Menentukan nilai mol begin mathsize 14px style N subscript 2 end style (begin mathsize 14px style italic A subscript r end style N = 14)

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space N subscript 2 space end cell equals cell space 2 cross times italic A subscript r space N end cell row blank equals cell space 2 cross times 14 end cell row blank equals cell space 28 space g space mol to the power of negative sign 1 end exponent end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript N subscript 2 end subscript space end cell equals cell space m over Mr end cell row blank equals cell space 7 over 28 end cell row blank equals cell space 0 comma 25 space mol end cell end table end style 

Maka volume gas begin mathsize 14px style C O subscript 2 end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript C O subscript 2 end subscript over V subscript C O subscript 2 end subscript space end cell equals cell space n subscript N subscript 2 end subscript over V subscript N subscript 2 end subscript end cell row cell fraction numerator 0 comma 1 over denominator V subscript C O subscript 2 end subscript end fraction space end cell equals cell space fraction numerator 0 comma 25 over denominator 10 end fraction end cell row cell V subscript C O subscript 2 end subscript space end cell equals cell space fraction numerator 1 over denominator 0 comma 25 end fraction end cell row cell V subscript C O subscript 2 end subscript space end cell equals cell space 4 space L end cell end table end style 

Jadi, jawaban yang benar adalah B.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 13 Maret 2021

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