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JAMB Chemistry 1983 Question 50

Question 50

Mole Concept and Stoichiometry

30 cm3 of 0.1 M Al(NO3)3 solution is reacted with 100cm3 of 0.15M of NaOH solution. Which reactant is in excess, and by how much?

  1. A.NaOH solution, by 70 cm3
  2. B.NaOH solution, by 60 cm3
  3. C.NaOH solution, by 40 cm3
  4. D.Al(NO3)3 solution, by 20 cm3
  5. E.Al(NO3)3 solution, by 10 cm3.
Show the answer and explanation

Answer: C (NaOH solution, by 40 cm3)

Explanation

Al3+ = 0.003 mol needs 3 × 0.003 = 0.009 mol OH− (= 60 cm3 of 0.15 M). NaOH supplied is 100 cm3, so 40 cm3 is in excess.