Organic Chemistry I

Chemistry Level 3

Propyne of 8 g 8\text{ g} reacts with 6.72 L 6.72\text{ L} of H X 2 \ce{H_2} , measured in STP with the presence of N i \ce{Ni} as a catalyst.

Given that all or quantity of Propyne and H X 2 \ce{H_2} are converted into products. What are the quantities of products?.

You are given that ArC=12, and ArH=1.

m M r \frac{m}{Mr} =n , V 22 , 4 \frac{V}{22,4} =n (stp)

0,3mol Propane and 0mol Propene 0,5mol Propane and 1,5mol Propene 0,2mol Propane and 0mol Propene 0,1mol Propane and 0,1mol Propene

Dimitris Kouroupos by dimitris kouroupos , 23, Greece

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1 solution

  • for propyne know has Mr=12+1+12+12=40 . n(propyne)= m M r \frac{m}{Mr} = 8 40 \frac{8}{40} =0.2mol propyne at the beginning
  • for H2 6,72L STP BECAUSE n(H2)= V 22 , 4 \frac{V}{22,4} = 6 , 72 22 , 4 \frac{6,72}{22,4} =0,3mol H2 at the beginning
CH3C≡CH +H2 ⇌ CH3CH=CH2
0,2 0,3 -
-0,2 -0,2 +0,2
- 0,1 0,2
  • with the end of the first reaction ,because we have excess Η2 and them hydrogen will react with the generated propene to react all the quantities of reactants
CH3CH=CH2 +H2 CH3CH2CH3
0,2 0,1 -
-0.1 -0,1 +0,1
0,1 - 0,1
  • IN THE END OF REACTION WE HAVE: 0,1mol Propene and 0,1mol Propane
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