Gravimetrik analizga kirish: haydash gravimetriyasi

111-bosqich: namunadagi massa oʻzgarishini hisoblash
Qizdirish jarayoni davomida namunaning oʻzgargan massasini hisoblash orqali chiqarib yuborilgan suvning miqdorini aniqlashimiz mumkin.
Suvning massasi H2O=Namunaning boshlangʻich massasi−Namunaning oxirgi massasi=9,51 g−9,14 g=0,37 g H2OSuvning massasi H2​O​=Namunaning boshlangʻich massasi−Namunaning oxirgi massasi=9,51g−9,14g=0,37g H2​O​
222-bosqich. Bugʻlangan suv massasini molga aylantirish
Mol nisbatidan foydalanib chiqarib yuborilgan suv miqdoridan BaCl2⋅2H2OBaCl2​⋅2H2​Ostart text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text miqdorini topish uchun dastlab bugʻlangan suv massasini molga aylantirishimiz kerak. Buning uchun suvning molekulyar massasidan 18,02 g/mol18,02g/mol18, comma, 02, start text, g, slash, m, o, l, end text foydalanamiz.
Suvning massasi=0,37 g H2O×1 mol H2O18,02 g H2O=2,05×10−2 mol H2OSuvning massasi=0,37g H2​O​×18,02g H2​O​1mol H2​O​=2,05×10−2mol H2​Ostart text, S, u, v, n, i, n, g, space, m, a, s, s, a, s, i, end text, equals, 0, comma, 37, start cancel, start text, g, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, times, start fraction, 1, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text, divided by, 18, comma, 02, start cancel, start text, g, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, end fraction, equals, 2, comma, 05, times, 10, start superscript, minus, 2, end superscript, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text
333-bosqich. Suvning miqdoridan BaCl2⋅2H2OBaCl2​⋅2H2​Ostart text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text miqdorini topish
Biz muvozanatda turgan reaksiyaning mol nisbatidan foydalanib suv molidan BaCl2⋅2H2OBaCl2​⋅2H2​Ostart text, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text molini topishimiz mumkin.
mol BaCl2⋅2H2O=2,05×10−2 mol H2O×1 mol BaCl2⋅2H2O2 mol H2O=1,03×10−2 mol BaCl2⋅2H2Omol BaCl2​⋅2H2​O=2,05×10−2mol H2​O​×2mol H2​O​1mol BaCl2​⋅2H2​O​=1,03×10−2mol BaCl2​⋅2H2​Ostart text, m, o, l, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, equals, 2, comma, 05, times, 10, start superscript, minus, 2, end superscript, start cancel, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, times, start fraction, 1, start text, m, o, l, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, divided by, 2, start cancel, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, O, end text, end cancel, end fraction, equals, 1, comma, 03, times, 10, start superscript, minus, 2, end superscript, start text, m, o, l, space, B, a, C, l, end text, start subscript, 2, end subscript, dot, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text