QUESTION 4 1 poin The Haber-Bosch process is a very important industrial process

Question

QUESTION 4 1 poin 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 QUESTION S 1 points If the reaction in the previous question produced 2.77 g NHs, what is the percent yield for this reaction? (Write your answer as a percent, but without the symbol: i.e. 12.3% would be entered as 12.3)

Explanation / Answer

4)

Molar mass of H2 = 2.016 g/mol

mass of H2 = 1.51 g

we have below equation to be used:

number of mol of H2,

n = mass of H2/molar mass of H2

=(1.51 g)/(2.016 g/mol)

= 0.749 mol

Molar mass of N2 = 28.02 g/mol

mass of N2 = 10.1 g

we have below equation to be used:

number of mol of N2,

n = mass of N2/molar mass of N2

=(10.1 g)/(28.02 g/mol)

= 0.3605 mol

we have the Balanced chemical equation as:

3 H2 + N2 ---> 2 NH3 +

3 mol of H2 reacts with 1 mol of N2

for 0.749 mol of H2, 0.2497 mol of N2 is required

But we have 0.3605 mol of N2

so, H2 is limiting reagent

we will use H2 in further calculation

Molar mass of NH3 = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

From balanced chemical reaction, we see that

when 3 mol of H2 reacts, 2 mol of NH3 is formed

mol of NH3 formed = (2/3)* moles of H2

= (2/3)*0.749

= 0.4993 mol

we have below equation to be used:

mass of NH3 = number of mol * molar mass

= 0.4993*17.03

= 8.51 g

Answer: 8.51 g

5)

% yield = actual mass*100/theoretical mass

= 2.77*100/8.51

= 32.6 %

Answer: = 32.6

Feel free to comment below if you have any doubts or if this answer do not work

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