Tasdiqlayman” Tabiiy fanlar fakulteti dekani

Masala yechimi:
Dastlab barcha reaksiya tenglamalarini yozib tenglashtirib olamiz:
2C₃H₇NH₂ + 10.5O₂ → 6CO₂ + 9H₂O + N₂
2CH₃NH₂ + 4.5O₂ → 2CO₂ + 5H₂O + N₂
2CH₄ + 2O₂ → 6CO₂ + 2H₂O
Reaksiyadan ko‘rinib turibdiki, CO₂ dastlabki moddalarning hammasidan hosil bo‘ladi, N₂ esa faqat propilamin va etilamindan hosil bo‘lar ekan. Bu masala 3 noma’lumli tenglama asosida yechiladi. Tenglamani tuzish uchun CO₂ va N₂ ning modda miqdorini topib olamiz:
n = _{ } = 0.45 mol aralashma
n = _{ } = 0.1 mol N₂
n = _{ } = 0.65 mol CO₂
2x 6x x
2C₃H₇NH₂ + 10.5O₂ → 6CO₂ + 9H₂O + N₂ 0.1 mol
2y 0.45 2y 0.65 y
2CH₃NH₂ + 4.5O₂ → 2CO₂ + 5H₂O + N₂
2z 6z
2CH₄ + 2O₂ → 6CO₂ + 2H₂O
2x + 2y +2z = 0.45
6x + 2y + z = 0.65 (sistema ayriladi)
4x = 0.2
x = _{ } = 0.05
2x=0.1 6x=0.3 x=0.05
2C₃H₇NH₂ + 10.5O₂ → 6CO₂ + 9H₂O + N₂ 0.1 mol
2y=0.1 2y=0.1 y=0.05
2CH₃NH₂ + 4.5O₂ → 2CO₂ + 5H₂O + N₂
2z 6z
2CH₄ + 2O₂ → 6CO₂ + 2H₂O

1. 
   – 0.05 = 0.05 (y N₂)
   

0.45 (das.aral) – 0.1 – 0.1 = 0.25 mol (CH₄)
_{ }= _{ } * 100% = 55.5 %
Javob: 55.5 % CH₄
4-masala.
Reaksiya unumi 75 % bo‘lganda 37.2 g anilindan tribrom anilin olishda ajralib chiqadigan gazni neytrallash uchun 10 % li natriy gidroksid eritmasidan (p = 1.2 g/ml) necha ml sarf bo‘ladi.


Masala yechimi:
Dastlab reaksiya tenglamasini yozib olamiz:
C₆H₅NH₂ + 3Br₂ → C₆H₂(Br₃)NH₂ + 3HBr
37.2 ----------- 100 %
x ------------- 75 %
x = _{ } = 27.9 g (C₆H₅NH₂)
Mr (C₆H₅NH₂) = 93
n = _{ } = 0.3 mol
0.3 x
C₆H₅NH₂ + 3Br₂ → C₆H₂(Br₃)NH₂ + 3HBr
1 3
x = _{ } = 0.9 mol (Br₂)
NaOH bilan faqat HBr reaskiyaga kirishishini inobatga olgan holda masalani ishlashda davom etamiz.
x 0.9
NaOH + HBr → NaBr + H₂O
1 1
x = _{ } = 0.9 mol (NaOH)
m = 0.9 * 40 = 36 g (NaOH)

36 g -------------- 10 %

x ---------------- 100 %
x = _{ } = 360 eritma
V = _{ } = _{ } = 300 ml
Javob: 300 ml.
5-masala.
Benzol, fenol va anilindan iborat 50 g aralashmaga ishqor eritmasi qo‘shilganda aralashma massasi 4.7 g ga kamaydi. Qolgan eritmaga HCl qo‘shib chayqatilganda 25.9 g tuz hosil bo‘ldi. Dastlabki aralashmadagi moddalarning massasini (g) mos ravishda aniqlang.


Masalani yechimi:
Dastlab reaksiya tenglamasini yozib olamiz, ishqor bilan faqat fenol ta’sirlashadi:
C₆H₅OH + NaOH → C₆H₅ONa + H₂O
Demak, masala shartida aytilganidek, ishqor ta’sirida 4.7 g ga kamygan. Bu massa fenolga to‘g‘ri keladi.
Qolgan aralashmaga HCl ta’sir ettirilsa, u bilan anilin reaksiyaga kirishib kompleks tuz hosil qiladi.
x 25.9
C₆H₅NH₂ + HCl → C₆H₅NH₂*HCl
93 129.5
x = _{ } = 18.6 g (anilin)
Benzol massasini topish uchun aralashmani umumiy massasidan fenol va anilinni massalarini ayiramiz:
50 – 4.7 – 18.6 = 26.7 g (benzol)