Question

1) 2C8H18(g)+25O2(g)→16CO2(g)+18H2O(g)

A)After the reaction, how much octane is left?

2)The Haber-Bosch process is a very important industrial process. In the Haber-Bosch process, hydrogen gas reacts with nitrogen gas to produce ammonia according to the equation 3H2(g)+N2(g)→2NH3(g)   The ammonia produced in the Haber-Bosch process has a wide range of uses, from fertilizer to pharmaceuticals. However, the production of ammonia is difficult, resulting in lower yields than those predicted from the chemical equation. 1.71 g H2 is allowed to react with 10.4 g N2, producing 1.98 g NH3.

A)   What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units.

   B)   What is the percent yield for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units.

Thank you so much!!!!

Answer 1

1) This is a balanced equation.. So after the reaction, all octane will react with oxygen.No octane will remain.Even this is theoritically true, practically may changes.

2)

(A) as the reaction,

3H₂ + N₂ ----> 2NH₃

Mole ratio H₂ : N₂ : NH₃ = 3:1:2

Initial H2 moles = 1.71 g / 2 gmol-1 = 0.855 mol

Initial N2 moles = 10.4 g / 28 gmol-1 = 0.371 mol

The limiting agent is N2 here.

According to the stoicheometry, 0.285 moles of N2 reacts with (3*0.285) moles of H2 and give (2*0.285) moles of NH3...

So produced NH3 moles = 0.57 mol

Theoritical yeild of NH3 = 0.57 mol * 17 gmol-1 = 9.69 g

(B) Percentage yeild = ( Experimental yeild / Theritical yeild ) * 100%

  Percentage yeild = ( 1.98 g / 9,69 g ) * 100%

= 20.4 %