Q100 · Practice questionComplex unfamiliar10 marks
QUESTION 100 (10 marks)
A miniature ammonia project generates hydrogen by electrolysis with a current of 2.00 A for 9648.5 s, at 100% efficiency. Hydrogen formation requires two electrons per molecule. The hydrogen then reacts with 0.0200 mol N₂ by . All ammonia produced is captured in 50.00 mL of 1.000 mol L⁻¹ HCl and the solution is made up to 100.0 mL. A 10.00 mL aliquot requires 15.00 mL of 0.1000 mol L⁻¹ NaOH to neutralise the remaining HCl. Assume this endpoint measures only the remaining strong acid. Use .
a) Calculate the moles of hydrogen generated. [3 marks]
b) Identify the limiting Haber reagent and calculate theoretical ammonia moles. [2 marks]
c) Use the acid-capture result to calculate actual ammonia moles. [3 marks]
d) Calculate percentage yield and one reason a real Haber reaction may not attain the stoichiometric maximum. [2 marks]
WORKED SOLUTION
10 marksPractice marking scheme
H₂ = 0.1000 mol; N₂ limiting; theoretical NH₃ = 0.0400 mol; actual NH₃ = 0.03500 mol; yield = 87.5%. Equilibrium or losses can prevent full yield.
C; , so H₂=0.1000 mol. N₂ needs only 0.0600 mol H₂, and yields at most 0.0400 mol NH₃. Aliquot remaining HCl= mol; whole solution=0.01500 mol. Initially HCl=0.05000 mol, so ammonia= mol by . Yield=0.03500/0.0400 ×100=87.5%.
- Calculates electrolysis charge. [1 marks]
- Calculates electron moles. [1 marks]
- Uses two electrons per H₂ to obtain 0.1000 mol. [1 marks]
- Identifies N₂ as limiting using 1:3 ratio. [1 marks]
- Calculates theoretical NH₃ = 0.0400 mol. [1 marks]
- Calculates remaining acid in the aliquot. [1 marks]
- Applies tenfold aliquot factor to residual acid. [1 marks]
- Subtracts residual from initial acid and obtains actual NH₃ = 0.03500 mol. [1 marks]
- Calculates yield 87.5%. [1 marks]
- Gives valid equilibrium/separation/loss reason. [1 marks]
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