Kristalbarium klorida yang massanya 1,22 gram dilarutkan ke dalam air, kemudian direaksikan dengan larutan natrium sulfat berlebih sesuai reaksi: Setelah disaring, endapan yang dihasilkan dikeringkan dan ditimbang. Ternyata, massa endapannya adalah 1,165 gram.
Dịch Vụ Hỗ Trợ Vay Tiền Nhanh 1s. BaCl2 + Na2SO4 => BaSO4 + 2 NaClmassa BaSO4 = 1,165 grMr BaSO4 = 233mol BaSO4 = 1,165 gr / 233 gr/mol = 0,005 molmol BaCl2 = mol BaSO4 koefisien sama = 0,005 molMr BaCl2 = 208massa BaCl2 = 0,005 mol x 208 gr/mol = 1,04 grammassa H2O = 1,22 gr - 1,04 gr = 0,18 grMr H2O = 18mol H2O = 0,18 gr / 18 gr/mol = 0,01 molx H2O = 0,01 / 0,005 = 2jumlah kristal BaCl2 = 2rumus kristalnya =
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kristal barium klorida yang massanya 1 22 gram