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Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
Consider the above reaction, the limiting reagent of the reaction and number of moles of NH3 formed, respectively are :

A

H2, 1.42 moles

B

H2, 0.71 moles

C

N2, 1.42 moles

D

N2, 0.71 moles

Answer

N2, 1.42 moles

Explanation

Solution

N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
28 g N2 reacts with 6 g H2 limiting reagent is N2
∴ Amount of NH3 formed on reacting 20 g N2 is,
=34×2028=\frac{34 \times 20}{28}
=24.28=24.28 g
=1.42= 1.42 moles
So, the correct option is (C): N2, 1.42 moles