{"id":308031,"date":"2022-11-15T16:22:56","date_gmt":"2022-11-15T09:22:56","guid":{"rendered":"https:\/\/quipperhome.wpcomstaging.com\/?p=308031"},"modified":"2023-02-06T10:13:46","modified_gmt":"2023-02-06T03:13:46","slug":"contoh-soal-stoikiometri","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/mapel\/matematika\/contoh-soal-stoikiometri\/","title":{"rendered":"Contoh Soal Stoikiometri dan Penyelesaiannya"},"content":{"rendered":"\n
\"\"<\/figure>\n\n\n\n

Hai Quipperian, di artikel sebelumnya, Quipper Blog sudah pernah membahas tentang pengertian stoikiometri, pereaksi pembatas, sampai senyawa hidrat. Apakah kamu masih ingat bagaimana langkah penyelesaian soal stoikiometri? Yuk, simak contoh soal berikut ini.<\/p>\n\n\n\n

Contoh Soal 1<\/h2>\n\n\n\n

Jika rumus molekul cuka adalah CH3<\/sub>COOH, rumus empirisnya adalah<\/p>\n\n\n\n

    \n\t
  1. CH2<\/sub>O<\/li>\n\t
  2. CHO<\/li>\n\t
  3. CHO2<\/sub><\/li>\n\t
  4. C2<\/sub>HO<\/li>\n\t
  5. CHO3<\/sub><\/li>\n<\/ol>\n\n\n\n

    Pembahasan:<\/p>\n\n\n\n

    Rumus empiris merupakan bentuk paling sederhana dari rumus molekul. Artinya, terdapat faktor pengali yang menghubungkan rumus empiris dan rumus molekul. Secara matematis, dirumuskan sebagai berikut.<\/p>\n\n\n\n

    Rumus molekul = (Rumus empiris)n<\/sub><\/p>\n\n\n\n

    \u2194CH3<\/sub>COOH = (Rumus empiris)n<\/sub><\/p>\n\n\n\n

    \u2194C2<\/sub>H4<\/sub>O2<\/sub> = (CH2<\/sub>O)2<\/sub><\/p>\n\n\n\n

    Jadi, rumus empiris cuka adalah CH2<\/sub>O<\/p>\n\n\n\n

    Jawaban: A<\/strong><\/p>\n\n\n\n

    Contoh Soal 2<\/h2>\n\n\n\n

    Rumus empiris fruktosa adalah CH2<\/sub>O. Jika massa molekul relatif fruktosa tersebut adalah 180, rumus molekulnya adalah \u2026. (Ar<\/em> H = 1, C = 12, dan O = 16)<\/p>\n\n\n\n

      \n\t
    1. C6<\/sub>H12<\/sub>O6<\/sub><\/li>\n\t
    2. C4<\/sub>H12<\/sub>O6<\/sub><\/li>\n\t
    3. C3<\/sub>H6<\/sub>O3<\/sub><\/li>\n\t
    4. C2<\/sub>H4<\/sub>O2<\/sub><\/li>\n\t
    5. C4<\/sub>H12<\/sub>O4<\/sub><\/li>\n<\/ol>\n\n\n\n

      Pembahasan:<\/p>\n\n\n\n

      Mula-mula, kamu harus menentukan massa molekul relatif (Mr) dari rumus empiris fruktosa, yaitu CH2<\/sub>O.<\/p>\n\n\n\n

      Mr = ( 1 \u00d7 Ar C) + (2 \u00d7 Ar H) + (1 \u00d7 Ar O)<\/p>\n\n\n\n

      \u2003 = 12 + (2×1) + 16<\/p>\n\n\n\n

      \u2003 = 30<\/p>\n\n\n\n

      Lalu, tentukan variabel pengali rumus empirisnya.<\/p>\n\n\n\n

      n \u00d7 Mr rumus empiris = Mr rumus molekul<\/p>\n\n\n\n

      n \u00d7 30 = 180<\/p>\n\n\n\n

      n = 180 : 30= 6<\/p>\n\n\n\n

      Dengan demikian, rumus molekulnya bisa dinyatakan sebagai berikut.<\/p>\n\n\n\n

      Rumus molekul = (rumus empiris)n<\/sub><\/p>\n\n\n\n

      \u2003\u2003\u2003\u2003\u2003\u2003\u2003= (CH2<\/sub>O)6<\/sub><\/p>\n\n\n\n

      \u2003\u2003\u2003\u2003\u2003\u2003\u2003= (C6<\/sub>H12<\/sub>O)6<\/sub><\/p>\n\n\n\n

      Jadi, rumus molekul fruktosa adalah (C6<\/sub>H12<\/sub>O)6<\/sub><\/p>\n\n\n\n

      Jawaban: A<\/strong><\/p>\n\n\n\n

      Contoh Soal 3<\/h2>\n\n\n\n

      Pada suatu percobaan, terdapat sampel freon yang terdiri dari 0,423 gr C, 2,503 gr Cl, dan 1,339 gr F. Rumus empiris freon tersebut adalah \u2026. (Ar F = 19, C = 12, dan Cl = 35,5)<\/p>\n\n\n\n

        \n\t
      1. CClF2<\/sub><\/li>\n\t
      2. CCl2<\/sub>F<\/li>\n\t
      3. C2<\/sub>Cl2<\/sub>F2<\/sub><\/li>\n\t
      4. CCl2<\/sub>F3<\/sub><\/li>\n\t
      5. CCl2<\/sub>F2<\/sub><\/li>\n<\/ol>\n\n\n\n

        Pembahasan:<\/p>\n\n\n\n

        Diketahui:<\/p>\n\n\n\n

        m C = 0,423 gr<\/p>\n\n\n\n

        m Cl = 2,503 gr<\/p>\n\n\n\n

        m F = 1,339 gr<\/p>\n\n\n\n

        Ditanya: rumus empiris freon = … ?<\/p>\n\n\n\n

        Jawab:<\/p>\n\n\n\n

        Untuk mencari rumus empiris freon, gunakan perbandingan mol antara ketiga unsur.<\/p>\n\n\n\n

        \"contoh<\/figure>\n\n\n\n

        Jadi, rumus empiris freon adalah CCl2<\/sub>F2<\/sub><\/p>\n\n\n\n

        Jawaban: E<\/strong><\/p>\n\n\n\n

        Contoh Soal 4<\/strong><\/h2>\n\n\n\n

        Perhatikan persamaan reaksi berikut<\/p>\n\n\n\n

        Fe2<\/sub>O3<\/sub> + 3CO \u2192 2Fe + 3CO2<\/sub><\/p>\n\n\n\n

        Dari reaksi tersebut, ternyata dihasilkan 224 gr besi. Massa besi (III) oksida yang bereaksi adalah \u2026. (Ar Fe = 56, C = 12, dan O = 16)<\/p>\n\n\n\n

          \n\t
        1. 560 gr<\/li>\n\t
        2. 300 gr<\/li>\n\t
        3. 240 gr<\/li>\n\t
        4. 320 gr<\/li>\n\t
        5. 450 gr<\/li>\n<\/ol>\n\n\n\n

          Pembahasan:<\/p>\n\n\n\n

          Diketahui:<\/p>\n\n\n\n

          m <\/em>Fe = 224 gr<\/p>\n\n\n\n

          Fe2<\/sub>O3<\/sub> + 3 CO \u2192 2 Fe + 3 CO2<\/sub><\/p>\n\n\n\n

          Ditanya: m Fe2<\/sub>O3<\/sub> = ?<\/p>\n\n\n\n

          Jawab:<\/p>\n\n\n\n

          Mula-mula, kamu harus menentukan mol dari Fe<\/p>\n\n\n\n

          \"kamu<\/figure>\n\n\n\n

          Lalu, tentukan mol besi (III) oksida melalui perbandingan koefisien.<\/p>\n\n\n\n

          \"tentukan<\/figure>\n\n\n\n

          Dengan demikian, diperoleh massa besi (III) yang bereaksi sebagai berikut.<\/p>\n\n\n\n

          \"diperoleh<\/figure>\n\n\n\n

          Jadi, massa besi (III) oksida yang bereaksi adalah 320 gr<\/p>\n\n\n\n

          Jawaban: D<\/strong><\/p>\n\n\n\n

          Contoh Soal 5<\/strong><\/h2>\n\n\n\n

          Seorang peneliti mereaksikan 8,1 gr aluminium dengan asam klorida. Reaksi tersebut menghasilkan aluminium klorida dan gas hidrogen. Pada suhu dan tekanan tertentu, 1 mol gas hidrogen memiliki volume 6 liter. Adapun persamaan reaksinya adalah sebagai berikut.<\/p>\n\n\n\n

          2Al + 6HCl \u2192 2AlCl3<\/sub> + 3H2<\/sub><\/p>\n\n\n\n

          Volume gas hidrogen yang dihasilkan pada reaksi tersebut adalah \u2026. (Ar<\/em> Al = 27)<\/p>\n\n\n\n

            \n\t
          1. 1,8 liter<\/li>\n\t
          2. 2,7 liter<\/li>\n\t
          3. 5,4 liter<\/li>\n\t
          4. 4,5 liter<\/li>\n\t
          5. 4,8 liter<\/li>\n<\/ol>\n\n\n\n

            Pembahasan:<\/p>\n\n\n\n

            Mula-mula, kamu harus menentukan jumlah mol aluminium (Al)<\/p>\n\n\n\n

            \"kamu<\/figure>\n\n\n\n

            Selanjutnya, tentukan jumlah mol H2<\/sub> melalui perbandingan koefisien seperti berikut.<\/p>\n\n\n\n

            \"tentukan<\/figure>\n\n\n\n

            Terakhir, tentukan volume gas hidrogen yang dihasilkan pada reaksi tersebut dengan perbandingan seperti berikut.<\/p>\n\n\n\n

            \"\"<\/figure>\n\n\n\n

            Jadi, volume gas hidrogen yang dihasilkan pada reaksi tersebut adalah 2,7 liter.<\/p>\n\n\n\n

            Jawaban: B<\/strong><\/p>\n\n\n\n

            Contoh Soal 6<\/strong><\/h2>\n\n\n\n

            Suatu reaksi berlangsung seperti persamaan berikut<\/p>\n\n\n\n

            X + 2Y \u2192 XY2<\/sub><\/p>\n\n\n\n

            Massa kedua unsur yang direaksikan sama, yaitu 150 gr. Persen massa unsur yang tersisa adalah \u2026. (Ar<\/em> X = 15, Y = 30)<\/p>\n\n\n\n

              \n\t
            1. 60%<\/li>\n\t
            2. 50%<\/li>\n\t
            3. 75%<\/li>\n\t
            4. 80%<\/li>\n\t
            5. 45%<\/li>\n<\/ol>\n\n\n\n

              Pembahasan:<\/p>\n\n\n\n

              Mula-mula, kamu harus menentukan jumlah mol unsur X dan Y.<\/p>\n\n\n\n

              \"kamu<\/figure>\n\n\n\n

              Lalu, kamu harus menentukan pereaksi pembatas pada reaksi tersebut. Pereaksi pembatas diperoleh dari hasil bagi unsur yang terkecil.<\/p>\n\n\n\n

              X = 10 \/ 1 = 10<\/p>\n\n\n\n

              Y = 5\/2 = 2,5<\/p>\n\n\n\n

              Dari perhitungan di atas, diketahui bahwa unsur Y akan habis bereaksi atau menjadi pereaksi pembatas. Perhatikan tabel berikut.<\/p>\n\n\n\n

              <\/th>x (mol)<\/th>+<\/th>2Y(mol)<\/th>\u2192<\/th>XY2<\/sub>(mol)<\/th><\/tr><\/thead>
              Mula-mula<\/td>10<\/td><\/td>5<\/td><\/td>–<\/td><\/tr>
              Reaksi<\/td>1\/2×5 = 2,5<\/td><\/td>5<\/td><\/td>1\/2×5 = 2,5<\/td><\/tr>
              Sisa<\/td>7,5<\/td><\/td>–<\/td><\/td>2,5<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

              Dengan demikian, persen massa unsur X yang tersisa adalah sebagai berikut.<\/p>\n\n\n\n

              \"persen<\/figure>\n\n\n\n

              Jadi, persen massa unsur yang tersisa adalah 75%<\/p>\n\n\n\n

              Jawaban: C<\/strong><\/p>\n\n\n\n

              Contoh Soal 7<\/h2>\n\n\n\n

              Reaksi pembakaran 32 gram gas metana berlangsung seperti reaksi berikut<\/p>\n\n\n\n

              CH4<\/sub> + 2O2<\/sub> \u2192 CO2<\/sub> + 2H2<\/sub>O<\/p>\n\n\n\n

              Jika massa oksigen yang digunakan pada pembakaran tersebut 160 gr, banyaknya mol H2<\/sub>O yang terbentuk adalah \u2026. (Ar C = 12, O = 16)<\/p>\n\n\n\n

                \n\t
              1. 4 mol<\/li>\n\t
              2. 3 mol<\/li>\n\t
              3. 2 mol<\/li>\n\t
              4. 2,5 mol<\/li>\n\t
              5. 5 mol<\/li>\n<\/ol>\n\n\n\n

                Pembahasan:<\/p>\n\n\n\n

                Mula-mula, kamu harus menentukan jumlah mol unsur metana (CH4<\/sub>) dan oksigen (O2<\/sub>).<\/p>\n\n\n\n

                \"kamu<\/figure>\n\n\n\n

                Lalu, kamu harus menentukan pereaksi pembatas pada reaksi tersebut. Pereaksi pembatas diperoleh dari hasil bagi unsur yang terkecil.<\/p>\n\n\n\n

                CH4<\/sub> = 2\/1 = 2<\/p>\n\n\n\n

                O2<\/sub> = 5\/2 = 2,5<\/p>\n\n\n\n

                Dari perhitungan di atas, diketahui bahwa gas metana akan habis bereaksi atau menjadi pereaksi pembatas. Perhatikan tabel berikut.<\/p>\n\n\n\n

                <\/th>CH4<\/sub>(mol)<\/th>2O2<\/sub>(mol)<\/th>\u2192<\/th>CO2<\/sub>(mol)<\/th>2H2<\/sub>O(mol)<\/th><\/tr><\/thead>
                Mula-mula<\/td>2<\/td>5<\/td><\/td>–<\/td>–<\/td><\/tr>
                Reaksi<\/td>2<\/td>4<\/td><\/td>2<\/td>4<\/td><\/tr>
                Sisa<\/td>–<\/td>1<\/td><\/td>2<\/td>4<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

                Jadi, banyaknya mol H2<\/sub>O yang terbentuk adalah 4 mol<\/p>\n\n\n\n

                Jawaban: A<\/strong><\/p>\n\n\n\n

                Contoh Soal 8<\/h2>\n\n\n\n

                Kristal CuSO4<\/sub> (Mr = 159,5)sebanyak 478,5 gr mampu menyerap 270 gram air. Akibatnya, terbentuk senyawa hidrat CuSO4<\/sub>.xH2<\/sub>O. Banyaknya molekul air yang terserap di dalam senyawa hidrat tersebut adalah<\/p>\n\n\n\n

                  \n\t
                1. 6<\/li>\n\t
                2. 4<\/li>\n\t
                3. 5<\/li>\n\t
                4. 3<\/li>\n\t
                5. 2<\/li>\n<\/ol>\n\n\n\n

                  Pembahasan:<\/p>\n\n\n\n

                  Mula-mula, kamu harus menentukan persamaan reaksinya seperti berikut<\/p>\n\n\n\n

                  CuSO4<\/sub> + xH2<\/sub>O \u2192 CuSO4<\/sub>.xH2<\/sub>O<\/p>\n\n\n\n

                  Selanjutnya, tentukan jumlah mol CuSO4<\/sub> dan H2<\/sub>O seperti berikut<\/p>\n\n\n\n

                  \"Selanjutnya,<\/figure>\n\n\n\n

                  Untuk mencari banyaknya molekul H2<\/sub>O pada senyawa hidrat CuSO4<\/sub>.xH2<\/sub>O, gunakan SUPER \u201cSolusi Quipper\u201d berikut.<\/p>\n\n\n\n

                  \"Untuk<\/figure>\n\n\n\n

                  Jadi, banyaknya molekul air pada senyawa hidrat tersebut adalah 5<\/p>\n\n\n\n

                  Jawaban: C<\/strong><\/p>\n\n\n\n

                  Contoh Soal 9<\/strong><\/h2>\n\n\n\n

                  Suatu senyawa hidrat MSO4<\/sub>.5H2<\/sub>O dikeringkan sehingga massanya berkurang 144 gram. Berapakah massa MSO4<\/sub> (Mr = 136) dalam senyawa hidrat tersebut?<\/p>\n\n\n\n

                  Pembahasan:<\/p>\n\n\n\n

                  Di soal tertulis bahwa massa senyawa hidrat akan berkurang 144 jika dikeringkan. Massa yang berkurang itu = massa air. Dengan demikian, kamu bisa menentukan jumlah mol airnya.<\/p>\n\n\n\n

                  \"tertulis<\/figure>\n\n\n\n

                  Lalu, tentukan jumlah mol MSO4 dengan cara SUPER \u201cSolusi Quipper\u201d berikut.<\/p>\n\n\n\n

                  \"tentukan<\/figure>\n\n\n\n

                  Terakhir, tentukan massa MSO4 dengan cara berikut.<\/p>\n\n\n\nm<\/em>MSO4<\/sub> = n MSO4<\/sub> x Mr<\/br>\n   =1,6x 136<\/br>\n   = 217,6 gram\n\n\n\n

                  Jadi, massa MSO4<\/sub> (Mr = 136) dalam senyawa hidrat tersebut adalah 217,6 gram.<\/p>\n\n\n\n

                  Contoh Soal 10<\/h2>\n\n\n\n

                  Unsur Z mampu membentuk senyawa dengan unsur O membentuk Z2O3. Jika massa unsur Z = 3,2 gr dan massa unsur O = 9,6 gr, berapakah massa atom relatif Z? (Ar O = 16)<\/p>\n\n\n\n

                  Pembahasan:<\/p>\n\n\n\n

                  2Z + 3O \u2192 Z2<\/sub>O3<\/sub><\/p>\n\n\n\n

                  Selanjutnya, tentukan jumlah mol O.<\/p>\n\n\n\n

                  \"Selanjutnya,<\/figure>\n\n\n\n

                  Lalu, tentukan mol unsur Z melalui perbandingan koefisien<\/p>\n\n\n\n

                  \"Lalu,<\/figure>\n\n\n\n

                  Dengan demikian, massa atom relatif Z bisa dinyatakan sebagai berikut<\/p>\n\n\n\n

                  \"Dengan<\/figure>\n\n\n\n

                  Jadi, massa atom relatif Z adalah 8<\/p>\n\n\n\n

                  Itulah pembahasan Quipper Blog kali ini. Semoga bermanfaat, ya. Untuk melihat contoh soal lainnya, yuk buruan gabung Quipper Video. Bersama Quipper Video, belajar jadi lebih mudah dan menyenangkan. Salam Quipper!<\/p>\n","protected":false},"excerpt":{"rendered":"

                  Hai Quipperian, di artikel sebelumnya, Quipper Blog sudah pernah membahas tentang pengertian stoikiometri, pereaksi pembatas, sampai senyawa hidrat. Apakah kamu masih ingat bagaimana langkah…<\/p>\n","protected":false},"author":156742495,"featured_media":308038,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_crdt_document":"","_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[679384865],"tags":[679384924],"ppma_author":[679386821,679386839],"class_list":["post-308031","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-matematika","tag-materi-kimia-kelas-10"],"acf":[],"yoast_head":"\nContoh Soal Stoikiometri dan Penyelesaiannya - Quipper Blog<\/title>\n<meta name=\"description\" content=\"Menguasai materi stoikiometri dengan mengerjekan contoh soal berikut jawaban dan langkah penyelsaiannya. 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