Question

12 g of 12C contains Avogadro’s number of carbon atoms.
a) Give the Avogadro’s number. (1)
b) The mass of 2 moles of ammonia gas is ………….
(i) 2 g (ii) 1.2 x 1022g (iii) 17 g (iv) 34g (1)
c) Calculate the volume of ammonia gas produced at STP when 140 g of nitrogen gas reacts with 30 g of
hydrogen gas.
(Atomic mass: N = 14u, H = 1 u) (2)

Answer 1

Answer :

a) 6.023 × 10²³

b) 34 g [Option IV)

Steps :

• Number of moles : Given mass/Molar mass
• 2 = Mass/17 [NH₃ = 14 + (3×1) = 17g]
• 2 × 17 = Mass
• Mass = 34 g

c) 224 litres

Steps :

1. No of moles of N₂ = 140/28 = 5 moles
2. No of moles of H₂ = 30/2 = 15 moles

N₂ + 3H₂ ➞ 2NH₃

• So from the above equation we see that 5 moles of N₂ would react with 15 moles of H₂ to produce 10 moles of NH₃
• Volume at STP = 10 × 22.4 = 224 litres

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Answer 2

Answer :

a) 6.023 × 10²³

b) 34 g [Option IV)

Steps :

• Number of moles : Given mass/Molar mass

• 2 = Mass/17 [NH₃ = 14 + (3×1) = 17g]

• 2 × 17 = Mass

• Mass = 34 g

c) 224 litres

Steps :

• No of moles of N₂ = 140/28 = 5 moles

• No of moles of H₂ = 30/2 = 15 moles

• N₂ + 3H₂ ➞ 2NH₃

So from the above equation we see that 5 moles of N₂ would react with 15 moles of H₂ to produce 10 moles of NH₃

Volume at STP = 10 × 22.4 = 224 litres

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