Isibonelo Sombuzo Wengxoxo Ngesixazululo se-Stoichiometry
I-Stoichiometry iyigatsha lekhemistri elifunda ubudlelwano obulinganiselwe phakathi kwama-reactants nemikhiqizo ekusabeleni kwamakhemikhali. I-stoichiometry yesixazululo icacile kakhulu, njengoba igxile ekusabeleni kwamakhemikhali okwenzeka ku-solvent, ngokuvamile emanzini. Kulesi sihloko, sizoxoxa ngezinkinga eziyisibonelo ezihlobene ne-stoichiometry yesixazululo ukuze siqonde kangcono lo mqondo.
Imiqondo Eyisisekelo Yesixazululo I-Stoichiometry
Ngaphambi kokuba singene emibuzweni yesibonelo, kuneminye imiqondo eyisisekelo okudingeka siyiqonde:
1. I-Molarity (M): Ukuhlushwa kwesisombululo kuvezwa nge-molarity. I-Molarity yinani lama-moles e-solute ngelitha lesisombululo (mol/L).
2. I-Mol: Iyunithi yenani lezinto lapho kukhona khona izinhlayiya ze-Avogadro (6.022 x 10²³).
3. Umthamo wesisombululo: Ngokuvamile ulinganiswa ngamalitha (L).
Ngalokhu kuqonda okuyisisekelo, sesikulungele ukuqhubekela ezinkingeni nezixazululo eziyisibonelo.
Isibonelo Umbuzo 1: Ukubala Ubuningi Besixazululo
Umbuzo:
Njengoba amagremu ama-5 e-NaCl (i-Sodium Chloride) encibilikiswa ku-250 mL wamanzi, ungakanani umkhawulo wesisombululo se-NaCl?
Isixazululo:
1. Bala inani lama-moles e-NaCl:
– Isisindo sama-molecule (BM) se-NaCl: Na (23) + Cl (35.5) = 58.5 g/mol
– Inani lama-moles = Ubuningi / BM
``
Inani lama-moles e-NaCl = 5 g / 58.5 g/mol ≈ 0.0855 mol
``
2. Guqula ivolumu yesisombululo kusuka ku-mL kuya ku-L:
``
Ivolumu yesisombululo = 250 mL = 0.250 L
``
3. Bala i-molarity (M):
``
I-Molarity (M) = Inani lama-moles / Umthamo wesisombululo
= 0.0855 mol / 0.250 L
≈ 0.342 M
``
Ngakho-ke, ubuningi besisombululo se-NaCl buyi-0.342 M.
Isibonelo Umbuzo 2: Ukusabela Kokungathathi-hlangothi
Umbuzo:
Yimuphi umthamo wesisombululo se-HCl esingu-0.5 M odingekayo ukuze kuncishiswe i-50 mL yesisombululo se-NaOH esingu-0.1 M?
Isixazululo:
Ukusabela okwenzekayo:
``
I-HCl + NaOH → I-NaCl + H₂O
``
Lokhu kusabela kwenzeka ngesilinganiso esingu-1:1.
1. Bala inani lama-moles e-NaOH:
``
Inani lama-moles e-NaOH = i-Molarity x ivolumu
= 0.1 M x 0.050 L
= 0.005 mol
``
2. Ngenxa yokuthi isilinganiso sokusabela singu-1:1, inani lama-moles e-HCl = inani lama-moles e-NaOH = 0.005 mol.
3. Bala ivolumu yesisombululo se-HCl esidingekayo:
``
Umthamo we-HCl = Inani lama-moles / i-Molarity
= 0.005 mol / 0.5 M
= 0.01 L (noma 10 mL)
``
Ngakho-ke, umthamo wesisombululo se-HCl esingu-0.5 M ungu-10 mL.
Isibonelo Umbuzo 3: I-Titration ye-Acid-Base
Umbuzo:
Isixazululo esingu-25 mL se-CH₃COOH (i-acetic acid) silinganiswa nesisombululo esingu-0.1 M NaOH endaweni yokulingana. Kudingeka isixazululo esingu-30 mL se-NaOH ukuze kufinyelelwe endaweni yokulingana. Bala ubuningi besisombululo se-CH₃COOH.
Isixazululo:
Ukusabela okwenzekayo:
``
CH₃COOH + NaOH → CH₃COONA + H₂O
``
Lokhu kusabela kwenzeka futhi ngesilinganiso esingu-1:1.
1. Bala inani lama-moles e-NaOH asetshenzisiwe:
``
Inani lama-moles e-NaOH = i-Molarity x ivolumu
= 0.1 M x 0.030 L
= 0.003 mol
``
2. Inani lama-moles e-CH₃COOH lifana nenani lama-moles e-NaOH asabela (isilinganiso 1:1):
``
Inani lama-moles e-CH₃COOH = 0.003 mol
``
3. Bala ubuningi be-CH₃COOH:
``
I-Molarity (M) = Inani lama-moles / Umthamo
= 0.003 mol / 0.025 L
= 0.12 M
``
Ngakho-ke, ubuningi besisombululo se-CH₃COOH bungu-0.12 M.
Isibonelo Umbuzo 4: Ukuxutshwa Kwesixazululo
Umbuzo:
Yimuphi umthamo wamanzi okumele ufakwe ukuze kuncibilikiswe i-100 mL yesisombululo se-H₂SO₄ esingu-1 M ukuze ukuhlushwa kwaso kube ngu-0.25 M?
Isixazululo:
1. Sebenzisa ifomula yokuxuba (M₁V₁ = M₂V₂):
– M₁ = 1 M
– V₁ = 100 mL
– M₂ = 0.25 M
2. Bala i-V₂:
``
M₁V₁ = M₂V₂
1 M x 100 mL = 0.25 M x V₂
V₂ = (1 x 100) / 0.25
= 400 ml
``
3. Bala umthamo wamanzi ozongezwa:
``
Umthamo wamanzi owengeziwe = V₂ – V₁
= 400 mL – 100 mL
= 300 ml
``
Ngakho-ke, kuyadingeka ukwengeza ama-300 mL amanzi ukuze kuncibilikiswe isisombululo se-H₂SO₄ kusuka ku-1 M kuya ku-0.25 M.
Isibonelo Umbuzo 5: Ukusabela Kwemvula
Umbuzo:
Mangaki amagremu e-AgCl precipitate akhiwa lapho i-100 mL yesisombululo se-0.1 M AgNO₃ ixutshwa ne-100 mL yesisombululo se-0.1 M NaCl?
Isixazululo:
Ukusabela okwenzekayo:
``
I-AgNO₃ + NaCl → I-AgCl (imvula) + NaNO₃
``
Lokhu kusabela kwenzeka ngesilinganiso esingu-1:1.
1. Bala inani lama-moles e-AgNO₃ kanye ne-NaCl:
``
Inani lama-moles e-AgNO₃ = Molarity x Volume
= 0.1 M x 0.100 L
= 0.01 mol
Inani lama-moles e-NaCl = i-Molarity x ivolumu
= 0.1 M x 0.100 L
= 0.01 mol
``
2. Ngenxa yokuthi isilinganiso sokusabela singu-1:1, inani lama-moles e-AgCl akhiwe = Inani lama-moles e-AgNO₃ noma i-NaCl:
``
Inani lama-moles e-AgCl = 0.01 mol
``
3. Bala isisindo se-AgCl:
– Isisindo sama-molecule se-AgCl = Ag (107.87) + Cl (35.45) = 143.32 g/mol
``
Isisindo se-AgCl = Inani lama-moles x BM
= 0.01 mol x 143.32 g/mol
= 1.4332g
``
Ngakho-ke, isisindo se-AgCl precipitate esakhiwe singamagremu angu-1.4332.
Isiphetho
Lesi sihloko sixoxe ngezinkinga eziningana zezibonelo kanye nezixazululo mayelana ne-stoichiometry yesisombululo, okuhlanganisa ukubala i-molarity, ukusabela kwe-neutralization, ama-titration e-acid-base, i-dilution, kanye nokwakheka kwe-precipitate. Ngokuqonda lezi zibonelo, singaqonda kangcono ukuthi umqondo we-stoichiometry yesisombululo usetshenziswa kanjani ezimweni zangempela. I-stoichiometry yesixazululo ayibalulekile nje kuphela esifundweni sekhemistri kodwa futhi inezinhlelo zokusebenza ezibanzi emikhakheni ehlukahlukene njengekhemisi, i-biochemistry, kanye nemboni yamakhemikhali.
