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STOIKIOMETRI

Berapa jumlah mol dari 2,4 gram Mg ?. STOIKIOMETRI. Jawab: (Ar Mg = 24). Berapa jumlah mol dari 1,2 gram MgSO 4 ?. 1 mol = 6,02 x 10 23 (Avogadro). Jawab: (Ar Mg = 24, S = 32, O = 16). Konsentrasi Larutan. 1,2 g MgSO 4 dilarutkan ke dalam 100 g air. Berapa molal konsentrasi larutannya ?.

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STOIKIOMETRI

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  1. Berapa jumlah mol dari 2,4 gram Mg ? STOIKIOMETRI Jawab: (Ar Mg = 24) • Berapa jumlah mol dari 1,2 gram MgSO4 ? 1 mol = 6,02 x 1023 (Avogadro) Jawab: (Ar Mg = 24, S = 32, O = 16)

  2. Konsentrasi Larutan • 1,2 g MgSO4 dilarutkan ke dalam 100 g air. Berapa molal konsentrasi larutannya ?

  3. 1,2 g MgSO4 dilarutkan ke dalam air sampai volumenya menjadi 100 mL. • Berapa molar konsentrasi larutannya ? • Berapa normal konsentrasi larutannya ? +2 MgSO4 (jadi n = 2)

  4. Pengenceran m1 . kgp1 = m2 . kgp2 mol1 = mol2 M1 . V1 = M2 . V2 N1 . V1 = N2 . V2 • 20 mL MgSO4 0,01 M ditambah air sampai 100 mL. Menjadi berapa konsentrasinya ? Jawab:

  5. Pembatas Reaksi • 2 mol C2H5OH direaksikan dengan 2 mol O2. Berapa mol CO2 yang dihasilkan? Jawab: C2H5OH(l) + 3O2 (g) → 2CO2 (g) + 3H2O(l) 2/1 2/3 sisa > habis R: (1/3)x2 mol 2 mol (2/3)x2 mol = 0,67 mol = 1,33 mol Sisa = 2 – 0,67 = 1,33 mol

  6. Persen Hasil • Pada percobaan pembakaran CH4 dihasilkan 0,5 gram air. Jika menurut perhitungan teori dihasilkan 0,6 gram air, maka berapa % hasilnya ? Jawab: Persen hasil

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