Question

In the raction


N2 + 3H2 -----> 2 NH3 ,

Determine the amount of NH3 formed if 2000g de - nitrogen reacts with 1000g dihydrogen.​

Answer 1

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=28.0 g

1 mol

Na(g)

+

=6.0 g

3 mol

3H

2(g)

→

=34.0 g

2 mol

2NH

3(g)

(i) 28.0 g of N

2

require 6.0 g of H

2

to produce

=34.0 g of NH

3

2.00×10

3

g of N

2

will produce

=

28

34

×2.00×10

3

g of NH

3

=2.43×10

3

g of NH

3

=2430 g NH

3

(ii) Yes, dihydrogen will remain unreacted to some extent

(iii) Amount of hydrogen that remains unreacted.

28.0 g of N

2

require 6.0 g of H

2

2.00 g×10

3

g of N

2

will require

=

28.0

6.0

×2.00×10

3

of H

2

=428.5 g of H

2

∴ Amount of hydrogen that remains unreacted

=[1.00×10

3

−428.5]

g

=571.5 g.

Answer 2

GIVEN

• N2 + H2 → 2NH3
• 2000 gram N2 reacts with 1000 gram hydrogen

TO FIND

• Amount of NH3 formed

SOLUTION

• N2 + 3H2 → 2NH3
• Molar mass of N2 = 14+14 = 28 g

Moles = Mass/Molar mass

• So, moles of nitrogen = 2000/28 = 71.42
• Molar mass of H2 = 2g
• Moles of Hydrogen = 1000/2 = 500.

Limiting reagent(LR) = Moles of reactant/ stoichiometric coefficient of reactant

For N2

• LR = 71.42/1 = 71.42

For H2

• LR = 500/3 = 166.6

Here, limiting reagent is N2

• 1 mole N2 gives 2 mole NH3
• 71.42 mole N2 gives 142.84 mole

Therefore, 142.84 mole of NH3 is formed.