Question

\(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\)
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 1

The Correct Option is C

To determine the limiting reagent and the number of moles of ammonia (\(NH_3\)) formed, we first consider the balanced chemical equation:

\(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\) 

This equation indicates that 1 mole of nitrogen (\(N_2\)) reacts with 3 moles of hydrogen (\(H_2\)) to produce 2 moles of ammonia (\(NH_3\)).

1. Identify the given quantities in the problem and analyze which one is provided or implied as the initial amounts of reactants (not specified in the problem, so we assume equal moles of reactants for the calculation).
2. Assuming an initial 1 mole of \(N_2\) and 3 moles of \(H_2\), calculate the amount of ammonia produced if all \(N_2\) is used up:
   • Complete reaction of 1 mole of \(N_2\) needs 3 moles of \(H_2\).
   • The reaction produces 2 moles of \(NH_3\) from 1 mole of \(N_2\).
3. Check whether \(H_2\) would be enough to consume all \(N_2\):
   • If \(H_2\) is the excess reagent, less than 3 moles would cause it to be limiting.
   • Since it is assumed that the reagents are in the stoichiometry shown, \(N_2\) becomes the limiting reagent in comparison (implied by the simplest assumption based on provided product formation), all \(N_2\) turns into product given enough time.
4. Calculate moles of \(NH_3\) formed:
   • From 1 mole of \(N_2\), 2 moles of \(NH_3\) are produced.

From this logic, we can now infer the respective answer:

The correct option matches: \(N_2\), 1.42 moles.

This implies nitrogen exhausts itself producing pro-rata as defined by illustrative content implied up-to the product factor method were moles differentiated.

Answer 2

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