Pertanyaan

Sebanyak 20 gram kalsium karbonat CaCO 3 ( M r = 100) bereaksi dengan asam klorida berlebihan menurut reaksi: CaCO 3 ( s ) + 2 HCl ( g ) → CaCl 2 ( a q ) + H 2 O ( l ) + CO 2 ( g ) . Pada suhu dan tekanan yang sama, 5 liter gas N 2 massanya 7 gram. Volume gas CO 2 yang dihasilkan adalah...( A r N = 14, C = 12, H = 1, O = 16)

Sebanyak 20 gram kalsium karbonat $CaCO_{3}$ ( $M_{r}$ = 100) bereaksi dengan asam klorida berlebihan menurut reaksi: $CaCO_{3} (s) + 2 HCl (g) \to CaCl_{2} (a q) + H_{2} O (l) + CO_{2} (g)$ . Pada suhu dan tekanan yang sama, 5 liter gas $N_{2}$ massanya 7 gram. Volume gas $CO_{2}$ yang dihasilkan adalah...( $A_{r}$ N = 14, C = 12, H = 1, O = 16)

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Klaim

A. Chandra

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

volume gas adalah 4 L.

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

adalah 4 L.

Pembahasan

Menentukan nilai mol : Menentukan volume : Jadi, volume gas adalah 4 L.

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 n subscript Ca C O subscript 3 end subscript end cell equals cell m over Mr end cell row blank equals cell 20 over 100 end cell row blank equals cell 0 comma 2 space mol end cell end table end style

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

Menentukan volume begin mathsize 14px style C O subscript 2 end style :

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript C O subscript 2 end subscript over V subscript C O subscript 2 end subscript end cell equals cell n subscript N subscript 2 end subscript over V subscript N subscript 2 end subscript end cell row cell fraction numerator 0 comma 2 over denominator V subscript C O subscript 2 end subscript end fraction end cell equals cell fraction numerator fraction numerator 7 over denominator left parenthesis 2 cross times 14 right parenthesis end fraction over denominator 5 end fraction end cell row cell fraction numerator 0 comma 2 over denominator V subscript C O subscript 2 end subscript end fraction end cell equals cell fraction numerator 0 comma 25 over denominator 5 end fraction end cell row cell V subscript C O subscript 2 end subscript end cell equals cell fraction numerator 0 comma 2 cross times 5 over denominator 0 comma 25 end fraction end cell row blank equals cell 4 space L end cell end table end style$

Jadi, volume gas adalah 4 L.