Isibonelo Sombuzo Wengxoxo Ye-Limiting Reagent
I-Pendahuluan
Kumakhemikhali, omunye wemibono eyisisekelo okuvame ukuxoxwa ngayo yi-reagent ekhawulelayo. I-reagent ekhawulelayo iyisakhi ekusabeleni kwamakhemikhali esizosetshenziswa kuqala, ngaleyo ndlela, sinquma inani lomkhiqizo ongakhiqizwa. Ukuqonda i-reagent ekhawulelayo kubalulekile ngoba kusisiza ukubikezela umphumela wokusabela kwamakhemikhali kanye nokusebenza kahle kokusetshenziswa kwamakhemikhali embonini noma elabhorethri.
Umqondo Wokunciphisa I-Reagent
Ngamazwi angokoqobo, i-reagent ekhawulelayo ingachazwa ngokufanisa. Ake sithi sifuna ukwenza amasangweji ngezithako ezimbili eziyinhloko: isinkwa nenyama. Uma sinezingcezu zesinkwa eziyi-10 kodwa sinezingcezu zenyama ezintathu kuphela, singenza amasangweji afinyelela kwayi-3. Kulokhu, inyama iyisithako esikhawulelayo, noma ngabe kusekhona isinkwa esisele.
Kumongo wekhemistri, umqondo ofanayo uyasebenza. Lapho ama-reactant amabili noma ngaphezulu esabela, elinye lawo lizophela kuqala, linciphise inani lomkhiqizo ongakhiwa. Lesi si-reactant sibe sesibizwa ngokuthi i-reagent ekhawulelayo.
Ukuhlonza i-Reactant Elinganiselwe Ekuphenduleni Kwamakhemikhali
Ukuze sithole i-reactant ekhawulelayo ekuphenduleni kwamakhemikhali, singasebenzisa izinyathelo ezilandelayo:
1. Bhala futhi ulinganisele i-equation yamakhemikhali yokusabela.
2. Bala inani lama-moles e-reactant ngayinye etholakalayo.
3. Sebenzisa i-stoichiometry (isilinganiso sama-mole) se-balanced equation ukuze uthole inani lomkhiqizo okhiqizwa yi-reactant ngayinye uma isisetshenziswe ngokuphelele.
4. Thola i-reactant ekhiqiza inani elincane kakhulu lomkhiqizo. Le reactant iyi-reactant ekhawulelayo.
Imibuzo Eyisibonelo Nengxoxo
umbuzo 1
Uma kubhekwa ukusabela okulandelayo:
\[ 2 \umbhalo{H}_2 + \umbhalo{O}_2 \umcibisholo ongakwesokudla 2 \umbhalo{H}_2\umbhalo{O} \]
Uma siqala ukusabela ngama-moles ama-5 e-H₂ nama-moles ama-2 e-O₂, nquma i-reagent ekhawulelayo nokuthi mangaki ama-moles e-H₂O azokhiqizwa.
Ingxoxo:
1. Bhala futhi ulinganisele izibalo zamakhemikhali:
\[ 2 \umbhalo{H}_2 + \umbhalo{O}_2 \umcibisholo ongakwesokudla 2 \umbhalo{H}_2\umbhalo{O} \]
2. Bala inani lama-moles e-reactant ngayinye:
– \( \umbhalo{H}_2 \) = 5 mol
– \( \text{O}_2 \) = 2 mol
3. Sebenzisa i-stoichiometry ukuthola inani lomkhiqizo ongakhiqizwa yi-reactant ngayinye:
– Kusukela ku-5 mol H₂:
\[ \text{Inani le-H₂O elingakhiwa} = 5 \text{ mol H}_2 \times \frac{2 \text{ mol H}_2\text{O}}{2 \text{ mol H}_2} = 5 \text{ mol H}_2\text{O} \]
– Kusukela kuma-moles amabili e-O₂:
\[ \text{Inani le-H₂O elingakhiwa} = 2 \text{ mol O}_2 \times \frac{2 \text{ mol H}_2\text{O}}{1 \text{ mol O}_2} = 4 \text{ mol H}_2\text{O} \]
4. I-reactant ekhiqiza inani elincane kakhulu lomkhiqizo yi-\( \text{O}_2 \). Ngakho-ke, i-\( \text{O}_2 \) iyi-reagent ekhawulelayo, futhi inani le-H₂O elingakhiqizwa lingama-moles angu-4.
umbuzo 2
Uma kubhekwa ukusabela okulandelayo:
\[ 4 \umbhalo{Al} + 3 \umbhalo{O}_2 \umcibisholo ongakwesokudla 2 \umbhalo{Al}_2\umbhalo{O}_3 \]
Uma siqala ngama-moles angu-8 e-Al kanye nama-moles angu-4 e-\( \text{O}_2 \), nquma i-reagent ekhawulelayo nokuthi zingaki ama-moles e-\( \text{Al}_2\text{O}_3 \) akhiqizwayo.
Ingxoxo:
1. Bhala futhi ulinganisele izibalo zamakhemikhali:
\[ 4 \umbhalo{Al} + 3 \umbhalo{O}_2 \umcibisholo ongakwesokudla 2 \umbhalo{Al}_2\umbhalo{O}_3 \]
2. Bala inani lama-moles e-reactant ngayinye:
– \( \text{Al} \) = 8 mol
– \( \text{O}_2 \) = 4 mol
3. Sebenzisa i-stoichiometry ukuthola inani lomkhiqizo ongakhiqizwa yi-reactant ngayinye:
– Kusukela kuma-moles angu-8 e-Al:
\[ \text{Amount } \text{Al}_2\text{O}_3 \text{ engakhiwa} = 8 \text{ mol Al} \times \frac{2 \text{ mol } \text{Al}_2\text{O}_3}{4 \text{ mol Al}} = 4 \text{ mol } \text{ Al}_2\text{O}_3 \]
– Kusukela kuma-moles angu-4 e-\( \text{O}_2 \):
\[ \text{Amount } \text{Al}_2\text{O}_3 \text{ engakhiwa} = 4 \text{ mol } \text{O}_2 \times \frac{2 \text{ mol } \text{Al}_2\text{O}_3}{3 \text{ mol } \text{O}_2} = 2.67 \text{ mol } \text{Al}_2\text{O}_3 \]
4. I-reactant ekhiqiza inani elincane kakhulu lomkhiqizo yi-\( \text{O}_2 \). Ngakho-ke, i-\( \text{O}_2 \) iyi-reagent ekhawulelayo, kanti inani le-\( \text{Al}_2\text{O}_3 \) elingakhiqizwa lingu-2.67 mol.
umbuzo 3
Uma kubhekwa ukusabela okulandelayo:
\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]
Uma siqala ngo-6 mol \( \text{N}_2 \) kanye no-18 mol \( \text{H}_2 \), nquma i-reagent ekhawulelayo nokuthi zingaki ama-moles ka-\( \text{NH}_3 \) akhiqizwayo.
Ingxoxo:
1. Bhala futhi ulinganisele ukusabela kwamakhemikhali:
\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]
2. Bala inani lama-moles e-reactant ngayinye:
– \( \umbhalo{N}_2 \) = 6 mol
– \( \umbhalo{H}_2 \) = 18 mol
3. Sebenzisa i-stoichiometry ukuthola inani lomkhiqizo ongakhiqizwa yi-reactant ngayinye:
– Kusukela kuma-moles ayi-6 \( \text{N}_2 \):
\[ \text{Amount } \text{NH}_3 \text{ engakhiwa} = 6 \text{ mol } \text{N}_2 \times \frac{2 \text{ mol } \text{NH}_3}{1 \text{ mol } \text{N}_2} = 12 \text{ mol } \text{NH}_3 \]
– Kusukela ku-18 mol \( \text{H}_2 \):
\[ \text{Amount } \text{NH}_3 \text{ engakhiwa} = 18 \text{ mol } \text{H}_2 \times \frac{2 \text{ mol } \text{NH}_3}{3 \text{ mol } \text{H}_2} = 12 \text{ mol } \text{NH}_3 \]
4. Inani eliphezulu lomkhiqizo okhiqizwayo lifana kuzo zombili izithasiselo, kodwa ngokwemfundiso inani elilinganiselwe lomkhiqizo lingu-\( \text{H}_2 \), ukuze \( \text{H}_2 \) libe yi-reactant ekhawulelayo.
Isiphetho
Umqondo wokunciphisa i-reagent ubalulekile ekuqondeni ukusabela kwamakhemikhali. Kusiza ukubikezela ukusebenza kahle kokusabela kanye nenani lomkhiqizo okhiqizwayo. Ezisetshenzisweni zezimboni, ukuqonda nokuhlonza i-reagents ezinciphisayo kubalulekile ekwenzeni ngcono ukusetshenziswa kwezinto zokusetshenziswa kanye nokunciphisa imfucuza.
Kulesi sihloko, kunezibonelo zezinkinga kanye nezingxoxo ezinikezwayo ukusiza abafundi baqonde izinyathelo zokuhlonza i-reagent ekhawulelayo. Ngokuzijwayeza kanye nokuqonda okuhle, lo mqondo ungasetshenziswa ezinhlobonhlobo eziningi zezindlela zamakhemikhali eziyinkimbinkimbi.
