Question

1. the reactant that runs out first and stops the reaction is called the limiting reactant. the one that is left over is called the excess reactant. in the third scenario in part 2 (with 6 n₂ and 6 h₂ used), identify the: limiting reactant: excess reactant: game go to the \game\ tab and try the levels to determine the correct amount of reactants, products, and leftovers in each question. there are 3 levels. start on level 1 and see how far you can get. level 1 score: 8/10 level 2 score: 7/10

Explanation:

Step1: Write the balanced chemical equation

The reaction between \(N_2\) and \(H_2\) to form \(NH_3\) is \(N_{2}+3H_{2}
ightarrow 2NH_{3}\).

Step2: Calculate the amount of reactants consumed based on stoichiometry

From the balanced equation, the mole ratio of \(N_2\) to \(H_2\) is \(1:3\). If we have \(n(H_{2}) = 6\) moles, the amount of \(N_{2}\) required for complete reaction of \(H_{2}\) is \(n(N_{2})=\frac{6}{3}=2\) moles. But we have \(n(N_{2}) = 6\) moles. If we consider the amount of \(N_{2}\), the amount of \(H_{2}\) required for complete reaction of \(N_{2}\) is \(n(H_{2})=3\times6 = 18\) moles. But we have only \(n(H_{2})=6\) moles.

Step3: Determine the limiting and excess reactant

Since \(H_{2}\) will be completely consumed first (as per the stoichiometry when calculating based on the available amounts), \(H_{2}\) is the limiting reactant. And \(N_{2}\) is in excess.

Answer:

limiting reactant: \(H_{2}\)
excess reactant: \(N_{2}\)