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Question 2

Ammonia is produced by Haber process according to the equation

N2(g) + 3H2(g)   →  2NH3(g)

If 20cm3 of Nitrogen were mixed with 90cm3 of Hydrogen under suitable conditions

(a) Which gas is in excess, and by how much?

Answer – \( \scriptsize \underset{1\;vol}{N_2} + \underset {3\; vol}{3H_{2(g)}} \; \rightarrow \; \underset{2\;vol}{2NH_{3(g)}} \)

Hydrogen; by 30cm3

(b) What will be the volume of Ammonia produced?

Answer – Volume before reaction

 :- \( \scriptsize \underset{20cm^3}{N_2} + \underset {90cm^3}{3H_{2(g)}} \; \rightarrow \; \underset{0cm^3}{2NH_{3(g)}} \)

After spark.

 

N2 3H2 →  2NH3(g)
20cm3 60cm3   40cm3

 

Volume of ammonia produced = 40cm3

Volume of excess/unused gas = (90 – 60)cm3 of H2

= 30cm3

(c) What is the total volume of the gas mixture at the end of the reaction?

Answer – Total volume of the gas mixture or resulting gases = 40cm3 of NH3 30cm3 of hydrogen

Question 3

(a) State Gay-Lussac’s Law of combining volume 

Answer – Gay-Lussac’s law of combining volume states that when gases react, they do so in volumes which bear a simple ratio with one another and to the volumes of the products if gaseous, provided that temperature and pressure remain constant.

(b) What volume of Hydrogen will be produced if 100cm3 of Ammonia is completely decomposed at constant temperature and pressure

           2NH3(g) →  N2(g) + 3H2(g)

Answer –

= \( \frac {100\; \times \;1}{2} \)

10cm3

Question 4

(a) State Avogadro’s Law 

Answer – Avogadro’s law states that equal volume of all gases at the same temperature and pressure contain the same number of molecules.

(b) 50cm3 of Hydrogen were sparked in a jar containing 50cm3 of Chlorine to produce Hydrogen Chloride gas at the same temperature and pressure 

(i) Write an equation for the reaction taking place 

Answer –

H2 + Cl2(g) →  2HCl(g)
1V   1V   2V

 

(ii) Calculate the composition of the gaseous mixture that would be obtained at the end of the reaction

Answer –

  H2 + Cl2(g) →  2HCl(g)
  1V   1V   2V
Volume before reaction 50cm3   50cm3   0cm3

 

Volume after reaction =  100cm3

Question 5

In the reaction: 2SO2(g) + O2(g)  → 2SO3(g)

2SO2(g) + O2(g) →  2SO3(g)
2V   1V   2V

(a) What volume of Sulphur(IV)Oxide, SO2, will combine with 50.0m3 of Oxygen gas?

Answer –

2cm3 will combine with 1cm3 of O2.

?cm3 of SO2 will combine with 50cm3 of O2

= \( \frac {50\; \times \;2}{1} \)

= 100cm3

Answer –

(b) Calculate the volume of Sulphur (VI) Oxide, SO3 produced.

Volume of SO3 produced = 2V = 100cm3

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