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e Content Question Completion Status QUESTION4 1 points Save 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) 2 NH3 (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 What is the maximum theoretical yield in grams if 1.51 g H2 is allowed to react with 10.1 g N2? QUESTION5 1 points Save If the reaction in the previous question produced 2.77 g NH3, what is the percent yield for this reaction? (Write your answer as a percent, but without the symbol, ie. 1 2.3% would be entered as 123)

Explanation / Answer

The reaction is

N2 + 3H2 -----------------> 2NH3

10.1/28=0.3607 1.51/2=0.755 0 initial moles

We need to find whichis limiting and whichis excess reagent.

0.3607/1 =0.3607 0.755/3=0.251

As the ratio is less for H2 , H2 is the limiting reagent andis consumed completely in the reaction.

N2 + 3H2 -----------------> 2NH3

10.1/28=0.3607 1.51/2=0.755 0 initial moles

0.3607-(0.755/3) 0 2 (0.755/3) after reaction

Thus the moles of NH3 produced = 2 (0.755/3) mol

mass of ammonia produced = 2 (0.755/3) mol x 17g/mol

=8.556 g

The maximum theoretical yield from the given quantities of H2 and N2 is 8,556 g

Q5) Actual yield = 2.77g

% yield = actual yield x 100/ theoretical yield

= 2.77x100 /8.556

= 32.37%

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