Pertanyaan

Pada pembakaran sempurna 1 liter (STP) suatu sampel gas alam dihasilkan kalor sebanyak 43,6 kJ. Jika gas alam tersebut hanya mengandung metana ( CH 4 ​ ) dan etana ( C 2 ​ H 6 ​ ), tentukan persen volume metana dalam campuran tersebut. ( △ H f ∘ ​ dari ; = − 85 kJ mol − 1 ; ; dan )

Pada pembakaran sempurna 1 liter (STP) suatu sampel gas alam dihasilkan kalor sebanyak 43,6 kJ. Jika gas alam tersebut hanya mengandung metana () dan etana (), tentukan persen volume metana dalam campuran tersebut.

( dari C H subscript 4equals minus sign 75 space kJ space mol to the power of negative sign 1 end exponentC subscript 2 H subscript 6C O subscript 2 equals minus sign 393 comma 5 space kJ space mol to the power of negative sign 1 end exponent; dan H subscript 2 O equals minus sign 242 space kJ space mol to the power of negative sign 1 end exponent)  space 

M. Dwiyanti

Master Teacher

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Jawaban terverifikasi

Jawaban

persen volume metana sebesar 71,9%.

persen volume metana sebesar 71,9%.

Pembahasan

Misalkan, Reaksi dan perubahan entalpi untuk 1 mol senyawa: Jadi, persen volume metana sebesar 71,9%.

Misalkan,

n space C H subscript 4 double bond x space mol n space C subscript 2 H subscript 6 equals fraction numerator 1 over denominator 22 comma 4 end fraction minus sign xmol  

Reaksi dan perubahan entalpi untuk 1 mol senyawa:

C H subscript 4 and 2 O subscript 2 yields C O subscript 2 and 2 H subscript 2 O 

table attributes columnalign right center left columnspacing 0px end attributes row cell increment H subscript c space C H subscript 4 end cell equals cell left parenthesis 1 cross times increment H subscript f degree space C O subscript 2 right parenthesis plus left parenthesis 2 cross times increment H subscript f degree space H subscript 2 O right parenthesis minus sign left parenthesis 1 cross times increment H subscript f degree space C H subscript 4 right parenthesis end cell row blank equals cell left parenthesis minus sign 393 comma 5 space kJ space mol to the power of negative sign 1 end exponent right parenthesis plus left parenthesis 2 cross times minus sign 242 space kJ space mol to the power of negative sign 1 end exponent right parenthesis minus sign left parenthesis minus sign 75 space kJ space mol to the power of negative sign 1 end exponent right parenthesis end cell row blank equals cell negative sign 802 comma 5 space kJ end cell end table 

C subscript 2 H subscript 6 plus 7 over 2 O subscript 2 yields 2 C O subscript 2 and 3 H subscript 2 O  

table attributes columnalign right center left columnspacing 0px end attributes row cell increment H subscript c space C subscript 2 H subscript 6 end cell equals cell left parenthesis 2 cross times increment H subscript f degree space C O subscript 2 right parenthesis plus left parenthesis 3 cross times increment H subscript f degree space H subscript 2 O right parenthesis minus sign left parenthesis 1 cross times increment H subscript f degree space C subscript 2 H subscript 6 right parenthesis end cell row blank equals cell left parenthesis 2 cross times minus sign 393 comma 5 space kJ space mol to the power of negative sign 1 end exponent right parenthesis plus left parenthesis 3 cross times minus sign 242 space kJ space mol to the power of negative sign 1 end exponent right parenthesis minus sign left parenthesis minus sign 85 space kJ space mol to the power of negative sign 1 end exponent right parenthesis end cell row blank equals cell negative sign 1.428 space kJ end cell end table  


table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell left parenthesis n cross times increment H space C H subscript 4 right parenthesis plus left parenthesis n cross times increment H space C subscript 2 H subscript 6 right parenthesis end cell row cell negative sign 43 comma 6 space kJ end cell equals cell left parenthesis x space mol cross times minus sign 802 comma 5 space kJ space mol to the power of negative sign 1 end exponent right parenthesis plus left parenthesis left parenthesis fraction numerator 1 over denominator 22 comma 4 end fraction minus sign x right parenthesis space mol cross times minus sign 1.428 space kJ space mol to the power of negative sign 1 end exponent right parenthesis end cell row cell negative sign 43 comma 6 end cell equals cell negative sign 802 comma 5 x plus 1.428 x minus sign 63 comma 75 end cell row cell 625 comma 5 x end cell equals cell 20 comma 15 end cell row x equals cell fraction numerator 20 comma 1 over denominator 625 comma 5 end fraction end cell row x equals cell 0 comma 0321 space mol end cell end table 


table attributes columnalign right center left columnspacing 0px end attributes row cell V space C H subscript 4 end cell equals cell x space mol cross times 22 comma 4 space L forward slash mol end cell row blank equals cell 0 comma 0321 space mol cross times 22 comma 4 space L forward slash mol end cell row blank equals cell 0 comma 719 space L end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell percent sign C H subscript 4 end cell equals cell fraction numerator V space C H subscript 4 over denominator V space sampel end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 0 comma 719 space L over denominator 1 space L end fraction cross times 100 percent sign end cell row blank equals cell 71 comma 9 percent sign end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell percent sign C subscript 2 H subscript 6 end cell equals cell 100 percent sign minus sign 71 comma 9 percent sign end cell row blank equals cell 28 comma 1 percent sign end cell end table 


Jadi, persen volume metana sebesar 71,9%.

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Diketahui data berikut. △ H f ∘ ​ CH 4 ​ ( g ) = − 74 , 8 kJmol − 1 △ H f ∘ ​ atomC ( g ) = + 715 kJmol − 1 △ H f ∘ ​ atomH ( g ) = + 218 kJmol − 1 D c − H ​ = 415 , 45 kJ mol − 1 Persamaan reak...

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