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JAMB Chemistry 1998 Question 4

Question 4

Gas Laws

C3H8(g) + 5O2(g) → 4H2O(g) + 3CO2(g). From the equation above, the volume of oxygen at s.t.p. required to burn 50cm3 of propane is

  1. A.250cm3
  2. B.150cm3
  3. C.100cm3
  4. D.50cm3
Show the answer and explanation

Answer: A (250cm3)

Explanation

By Gay-Lussac's law, gas volumes react in the ratio of the coefficients: 1 volume of propane needs 5 volumes of oxygen, so 50 × 5 = 250 cm3.