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Pertanyaan

Gas metanaC H subscript 4 (Mr 16) sebanyak 64 gram dapat bereaksi dengan gas C l subscript 2 berlebih menghasilkan 50,5 gram gas C H subscript 3 C l (Mr 50,5) 170 gram gas C H subscript 2 C l subscript 2 (Mr 85) dan X gram gas C H C l subscript 3 (Mr 120). Setelah dihitung dengan cermat dan teliti, maka massa gas C H C l subscript 3 adalah.....

  1. 6 gram

  2. 12 gram

  3. 30 gram

  4. 60 gram

  5. 120 gram

A. Tiara

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

Jawaban

massa C H C l subscript 3 atau x adalah 120 gram

Pembahasan

Reaksi yang terjadi adalah

C H subscript 4 space space space space space space space space plus space space space space space C l subscript 2 space space space rightwards arrow space space space space C H subscript 3 C l space space space space plus space space C H subscript 2 C l subscript 2 space space plus space space space space C H C l subscript 3  64 space g r a m 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 50 comma 5 space g r a m space space space space 170 space g r a m space space space space space space x space g r a m space

Konsep dalam percobaan ini adalah, massa C yang terlibat dalam reaksi akan selalu sama. Sehingga C dalam metana akan terbagi pada 3 senyawa produk

 

Massa C reaktan bold left parenthesis bold italic d bold italic a bold italic l bold italic a bold italic m bold space bold italic C bold italic H subscript bold 4 bold right parenthesis

M a s s a space C equals fraction numerator j u m l a h space C space cross times space A r space C over denominator M r space C H subscript 4 end fraction cross times m a s s a space C H subscript 4  M a s s a space C equals fraction numerator 1 cross times 12 over denominator 16 end fraction cross times 64 space g r a m space  M a s s a space C equals 48 space g r a m space

 

Massa C produk

M a s s a space C space d a l a m space C H subscript 3 C l equals fraction numerator j u m l a h space C space cross times space A r space C over denominator M r space C H subscript 3 C l end fraction cross times m a s s a space C H subscript 3 C l  M a s s a space C space d a l a m space C H subscript 3 C l equals fraction numerator 1 cross times 12 over denominator 50 comma 5 end fraction cross times 50 comma 5 space g r a m space  M a s s a space C space d a l a m space C H subscript 3 C l equals bold 12 bold space bold italic g bold italic r bold italic a bold italic m bold space    M a s s a space C space d a l a m space C H subscript 2 C l subscript 2 equals fraction numerator j u m l a h space C space cross times space A r space C over denominator M r space C H subscript 2 C l subscript 2 end fraction cross times m a s s a space C H subscript 2 C l subscript 2  M a s s a space C space d a l a m space C H subscript 2 C l subscript 2 equals fraction numerator 1 cross times 12 over denominator 85 end fraction cross times 170 space g r a m space  M a s s a space C space d a l a m space C H subscript 2 C l subscript 2 equals bold 24 bold space bold italic g bold italic r bold italic a bold italic m bold space

 

M a s s a space C thin space r e a k tan space equals space M a s s a space C space p r o d u k  M a s a space C thin space left parenthesis C H subscript 4 right parenthesis space equals space m a s s a space C thin space left parenthesis C H subscript 3 C l right parenthesis space plus thin space m a s s a space C space left parenthesis C H subscript 2 C l subscript 2 right parenthesis thin space plus thin space m a s s a space C space left parenthesis C H C l subscript 3 right parenthesis  48 space g r a m space equals space 12 space g r a m space plus space 12 space g r a m space plus space 24 space g r a m space plus space m a s s a space C space left parenthesis C H C l subscript 3 right parenthesis  m a s s a space C space left parenthesis C H C l subscript 3 right parenthesis equals space bold 12 bold space bold italic g bold italic r bold italic a bold italic m bold space

 

Menentukan massa C H C l subscript 3

M a s s a space C H C l subscript 3 equals fraction numerator M r space space C H C l subscript 3 over denominator j u m l a h space C space cross times space A r space C end fraction cross times m a s s a space C thin space d a l a m space space C H C l subscript 3  M a s s a space C H C l subscript 3 equals fraction numerator 120 over denominator 1 cross times 12 end fraction cross times 12 space g r a m space  M a s s a space C H C l subscript 3 equals 120 space g r a m space

 

Jadi massa C H C l subscript 3 atau x adalah 120 gram

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