reaction enthalpy Calculation 9 (thermo-21) In One of them Of which: (Diameters 6 cm, LOther 72 cm) will be a few Drops Containing 85% or more by weight of petroleum (1/4 Milligrams) given and with This Air-pentan mixture Inflamed. a) Calculate The reaction enthalpy. b) Experimental shows It 's The Commission, that to a lot Containing 85% or more by weight of petroleum No better Explosion supplies, This Air-pentan mixture should be So what do you mean? Optimally be. Calculate The maximum Quantity to Containing 85% or more by weight of petroleum. Background: It 's wonderful . Try it . for The Teaching. Notes: Unknown Size – molecules self to erude.

  Aufgabe: thermo-21    

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In One of them Of which: (Diameters 6 cm, LOther 72 cm) will be a few Drops Containing 85% or more by weight of petroleum (1/4 Milligrams) given and with This Air-pentan mixture Inflamed.
a) Calculate The reaction enthalpy.
b) Experimental shows It 's The Commission, that to a lot Containing 85% or more by weight of petroleum No better Explosion supplies, This Air-pentan mixture should be So what do you mean? Optimally be. Calculate The maximum Quantity to Containing 85% or more by weight of petroleum.
Background: It 's wonderful . Try it . for The Teaching.
Notes: Unknown Size – molecules self to erude.

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Lösung:

Notes: Containing 85% or more by weight of petroleum C₅H₁₂, M(C₅H₁₂), Dichte(C₅H₁₂) = 0.63 g/ml

1. Assumptions:
       Containing 85% or more by weight of petroleum C₅H₁₂, M(C₅H₁₂) = 72 g/mol
       Dichte(C₅H₁₂) = 0.626 g/ml
       dHf0(C₅H12) = -147 kJ/mol
       All of them Participants Substances Gaseous
       experiment In the case of 25℃ (293 K)
2. a) reaction equation:
       1·C₅H₁₂ (g) + 8·O₂ (g) ⇄ 5·CO₂(g) + 6·H₂O(g)
3. ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
   ΔHoR = (5· ΔHof(CO₂,g) + 6· ΔHof(H₂O, g)) – ( ΔHof(C₅H₁₂, g) + 8· ΔHof(O₂,g))
   ΔHoR = (5·-393 + 6·-242) – (-147 + 8·0) = -3270 kJ/mol
4. real of which: Energy:
   Dichte(C₅H₁₂) = 0.626 g/ml
   This means that 1/4 Milligrams: 0.15 Grams (=0.0021 mol)
   E = – 6.81 KJ
5. b) calculation maximum Quantity Containing 85% or more by weight of petroleum
6. Volumes Of which::
       V = pi · r² · h = 3.141 · 0.03² · 0.72 = 0.00204 m³
       V about. 2 Litre
7. Ideal Gas 1 mol In the case of …
       0℃ (273 K) 22.4 l
       25℃ (298 K): 24 Litre
       2 Litre (The Commission shall adopt implementing acts.. Of which:) So what do you mean?: 1/12 mol
8. Air exists to 20% O₂, 80% N₂
       This means that 1/60 mol O₂ in The Of which:
       Reason reaction equation 8· less C₅H₁₂ as O₂, d.h. So what do you mean? 1/480 mol C₅H₁₂
       With M(C₅H₁₂) = 72 g/mol following: 0.15 Grams C₅H₁₂
       With Dichte(C₅H₁₂) = 0.63 g/ml following: So what do you mean? about. 1/4 Milligrams C₅H₁₂

Es will be So what do you mean? 6.81 KJ Released. The maximum Quantity to Containing 85% or more by weight of petroleum is about. 1/4 Milligrams

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