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Related Topic: The Mole: Formulae and Chemical Equations

Related Subtopic: Combining volumes of gases



Questions 2.14

  1. In the same manner shown above, rewrite Equation 2 in terms of volume.
  2. Determine the volume of oxygen gas required to react completely with 12 dm3 of nitrogen (II) oxide (NO).
  3. When a sample of methane gas (CH4) was burnt completely in air, 8.0 dm3 of carbon (IV) oxide, measured at s.t.p was produced. Determine (a) the mass of methane burnt and (b) mass in g of steam produced (Molar gas volume at s.t.p. = 22.4 dm3; H = 1; O = 16).
  4. A mixture of nitrogen (N2) and oxygen (O2) gases was sparked then allowed to cool to room temperature and pressure. The resulting mixture consisted of 100 cm3 of nitrogen (II) oxide and 60 cm3 of excess nitrogen gas. What were the initial volumes of oxygen and nitrogen?
  5. A certain volume of ammonia gas combined directly with 80 cm3 of hydrogen chloride, measured at r.t.p. What was (a) the volume of ammonia used, and (b) mass of the product?
  6. Explain why it would not be realistic to ask for the volume of product in Question 5?
  7. A mixture of equal volumes of hydrogen and chlorine gases was reacted in a rigid container, and the product allowed to cool to the original temperature of the mixture. Explain why there was no overall change in pressure.



Answers to Questions 2.14


  1. 2volCO(g) + 1volO2(g) = 2volCO2(g)
  2. 2NO(g) + O2(g) = 2NO2(g)
    2vol of NO2 require 1 vol of O2
    Therefore 12 dm3 require (12/2) x 1 = 6 dm3

  3. CH4(g) + O2(g) = H2O(g) + CO2(g)
    (a) 1vol of CO2 is produced by 1vol of CH4
    Therefore 5.6 dm3 is produced by 5.6/1) x 1 = 5.6 dm3
    (b) 22.4 dm3 of CH4 produces 2(1x2 + 16) = 36 g of H2O
    Therefore 5.6 dm3 produces 5.6/22.4) x 36 = 9 g.

  4. N2(g) + O2(g) = 2NO(g)
    2vol of NO are produced by 1vol of O2.
    Therefore 100 cm3 is produced by (100/2) x 1 = 50 cm3 of O2 (used)
    So initial volume of N2 = 50 cm3 (same as for O2 from the equation)
    Initial volume of O2 = 50 + 60 = 110 cm3.

  5. NH3(g) + HCl(g) = NH4Cl(s)
    Volume of NH3 used = 80 cm3 (same as for HCl from the equation)
    24 000 cm3 of HCl produces 14 + (1x4) + 35.5 = 53.5 g of NH4Cl
    Therefore 80 cm3 produces (80/24 000) x 53.5 = 1.783 g

  6. The product is solid; so Gay Lussac,s law does not apply to it. Moreover, we would need to know its density.
  7. H2(g) + Cl2(g) = 2HCl(g)
    From the equation, total volume (or moles) of reacting gases equals volume (or moles) of product (HCl(g)); so they exert the same pressure.