Roboguru

Pertanyaan

Pada pembakaran sempurna 40 gram tembaga dibutuhkan 35 gram gas oksigen dengan persamaan kimia berikut. 

begin mathsize 14px style Cu open parentheses italic s close parentheses and O subscript 2 open parentheses italic g close parentheses yields Cu O open parentheses italic s close parentheses end style 

Berapa gram massa tembaga (II) oksida yang dihasilkan pada pembakaran tersebut.space space

(begin mathsize 14px style A subscript r space Cu equals 63 comma 5 semicolon space O equals 16 end style)

J. Siregar

Master Teacher

Mahasiswa/Alumni Universitas Negeri Medan

Jawaban terverifikasi

Pembahasan

Penentuan massa tembaga (II) oksida sesuai langkah-langkah penyelesaian berikut.

  • Setarakan persamaan reaksi.

begin mathsize 14px style 2 Cu open parentheses italic s close parentheses and O subscript 2 open parentheses italic g close parentheses yields 2 Cu O open parentheses italic s close parentheses end style 

  • Hitung massa molar (dalam satuan begin mathsize 14px style g space mol to the power of negative sign 1 end exponent end style) reaktan. Massa molar (begin mathsize 14px style M subscript m end style) Cu sama dengan massa atom relatif (begin mathsize 14px style A subscript r end style), sedangkan massa molar (begin mathsize 14px style M subscript m end stylebegin mathsize 14px style O subscript 2 end style sama dengan massa molekul relatifnya (begin mathsize 14px style M subscript r end style).

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript m space Cu end cell equals cell A subscript r space Cu end cell row cell M subscript m space Cu end cell equals cell 63 comma 5 space g space mol to the power of negative sign 1 end exponent end cell row blank blank blank row cell M subscript m space O subscript 2 end cell equals cell M subscript r space O subscript 2 end cell row cell M subscript m space O subscript 2 end cell equals cell 2 cross times A subscript r space O end cell row cell M subscript m space O subscript 2 end cell equals cell 2 cross times 16 end cell row cell M subscript m space O subscript 2 end cell equals cell 32 space g space mol to the power of negative sign 1 end exponent end cell end table end style 
 

  • Hitung mol masing-masing reaktan.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n space Cu end cell equals cell m over M subscript m end cell row cell n space Cu end cell equals cell fraction numerator 40 space g over denominator 63 comma 5 space g space mol to the power of negative sign 1 end exponent end fraction end cell row cell n space Cu end cell equals cell 0 comma 63 space mol end cell row blank blank blank row cell n space O subscript 2 end cell equals cell m over M subscript m end cell row cell n space Cu end cell equals cell fraction numerator 35 space g over denominator 32 space g space mol to the power of negative sign 1 end exponent end fraction end cell row cell n space Cu end cell equals cell 1 comma 09 space mol end cell end table end style 

  • Tentukan pereaksi pembatas. Pereaksi pembatas ditentukan berdasarkan hasil begin mathsize 14px style mol over koefisien end style yang paling kecil.

begin mathsize 14px style fraction numerator n space Cu over denominator koefisien end fraction colon fraction numerator n space O subscript 2 over denominator koefisien end fraction fraction numerator 0 comma 63 space mol over denominator 2 end fraction colon fraction numerator 1 comma 09 space mol over denominator 1 end fraction bold 0 bold comma bold 315 bold space bold mol bold space bold colon bold space 1 comma 09 end style

Pereaksi pembatas sesuai perbandingan diatas adalah tembaga (Cu).

  • Hitung mol CuO berdasarkan data pereaksi pembatas.

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n space Cu O end cell equals cell fraction numerator koefisien space Cu O over denominator koefisien space Cu O end fraction cross times space n space Cu end cell row cell n space Cu O end cell equals cell 2 over 2 cross times space 0 comma 63 space mol end cell row cell n space Cu O end cell equals cell 0 comma 63 space mol end cell end table end style 

  • Hitung massa CuO.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript m space Cu O end cell equals cell 79 comma 5 space g space mol to the power of negative sign 1 end exponent end cell row blank blank blank row cell m space Cu O end cell equals cell n cross times M subscript m end cell row cell m space Cu O end cell equals cell 0 comma 63 space mol cross times 79 comma 5 space g space mol to the power of negative sign 1 end exponent end cell row cell m space Cu O end cell equals cell 50 comma 085 space g end cell end table end style 


Jadi, massa tembaga (II) oksida yang dihasilkan adalah 50,085 gram.

13

0.0 (0 rating)

Pertanyaan serupa

Sebanyak 20 gram CaCO3​ dilarutkan dalam 1 liter larutan asam klorida 0,2 M menurut persamaan reaksi: CaCO3​(s)+2HCl(aq)→CaCl2​(aq)+CO2​(g)+H2​O(l)  Tentukan massa CaCl2​ yang terbentuk (Ar​ Ca = 40...

100

0.0

Jawaban terverifikasi

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

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia