Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia 2 NH3 (g) N,(g) + 3 Н,(g) — Assume 0.190 mol N, and 0.604 mol H, are present initially After complete reaction, how many moles of ammonia are produced? NH3 mol How many moles of H, remain? Н: mol How many moles of N, remain? N2 mol What is the limiting reactant? Onitrogen O hydrogen


NH​​​​​3 : 0.380 mol (produced)

H​​​​​​2 : 0.034 mol (remains)

N​​​​​​2 : 0 mol (remains)

Nitrogen is the limiting reagent.

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In the whole process 0.380 mol of ammoniaproduced.

All the 0.190 mol of N​​​​​​2 is consumed and0 mol ofN​​​​​​2 remains.

0.570 mol of H​​​​​​2 is consumed in the reaction so,the amount of remaining H​​​​​​2 is, 0.604 - 0.570 =0.034 mol ofH​​​​​​2 remains.

Now the limiting reagent,

Limiting reagent is the reagent in the reaction which getscompletely consumed in the process. In this reactionN​​​​​​2 is the reactant that hets completely consumedafter the complete consumption of N​​​​​2 no furtherproduction of ammonia can take place. Thereby, the amount ofN​​​​​​2 limits the production of ammonia.

So, the limiting reagent is N​​​​​​2.

N₂ (g) + 3 H₂(g) 1 mal 3 ona 2NH₃ Cg) 2 mol
As we can see in the above reeaction, 1 mal of Ne reacts with 3 mal of Hy to preoduce 2 mal of NH₃.
In the question it is given that, 0. 190 mol of Na & 0.604 mol of H₂ arce preesent initially. From the equation of the reeaction, we know that, 1 mol of No reacts with 3 mol of H₂ to produce 2 mol of NH₃. (0.190) mol of Ny will react with (0.190x3=0.570 mol of He to preoduce, (6.19082=0.380) mol of NH₃. so, 0.190 mod of Na reacts with 0.570 mol of te produce 0.380 mal of NH3 to
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