Mukarram Qodirova Mavzu bayoni

Sana : 05.05.2020

Mavzu : Uglevodorodlar bobiga doir masalalar yechish

O‘qituvchi: Lobar Boboqulova

Mavzu bayoni:

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  Masala: 2,8 g alkenni to’la gidrogenlash uchun 0, 896 litr vodorod sarf bo’lsa, alkenni toping
  

2,8g 0,896 l

C_nH_2n + H₂ → C_nH_{2n + 2} 14n = (2,8*22,4)/0,896 = 70 n = 70/14 =5 C₅H₁₀

Javob: C₅H₁₀ penten

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  Masala: 20 litr noma’lum alkinni to’liq yondirish uchun 170 litr kislorod sarflandi. Noma’lum alkinni toping
  

20 l 170 l

C_nH_2n-2 + (1,5n – 0,5)O₂ → nCO ₂ + (n+1)H₂O 1,5n -0,5 = 170/20 n = 6 C₆H₁₀

Javob: C₆H₁₀ geksin

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  Masala: Noma’lum alkanning 6 litrini to’liq yoqish uchun 32,5 litr ozon-kislorod aralashmasi sarflandi. Aralashmadagi kislorodning hajmiy ulushi 60% ga teng. Noma’lum alkanni toping.
  

6 l 39 l

C_nH_2n+2 + (1,5n + 0,5)O₂ →nCO ₂ + (n+1)H₂O

V(O₂) = (32,5*60)/100 = 19,5 l V(O₃) = 32,5 – 19,5 = 13 l

2O₃ → 3O₂ 2mol (O₃) — 3 mol (O₂)

13 l (O₃) — x = (13*3)/2 = 19,5 litr

V(O₂) = 19,5 + 19,5 = 39 litr 1,5n + 0,5 = 6,5 n = 4 C₄H₁₀

Javob: C₄H₁₀ butan

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  Masala: Quyidagi o’zgarishlarni amalga oshirish uchun zarur bo’lgan reaksiyalarni yozing.
  

CaCO₃ → CaO → CaC₂ → C₂H₂ → CH₃COH →CH₃COOH

Uyga vazifa: 0,25 mol alkan tarkibida 12,04 * 10²³ ta vodorod atomi bo’lsa, ushbu alkanni toping.

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