NH3 : 0.380 mol (produced)
H2 : 0.034 mol (remains)
N2 : 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 N2 is consumed and0 mol ofN2 remains.
0.570 mol of H2 is consumed in the reaction so,the amount of remaining H2 is, 0.604 - 0.570 =0.034 mol ofH2 remains.
Now the limiting reagent,
Limiting reagent is the reagent in the reaction which getscompletely consumed in the process. In this reactionN2 is the reactant that hets completely consumedafter the complete consumption of N2 no furtherproduction of ammonia can take place. Thereby, the amount ofN2 limits the production of ammonia.
So, the limiting reagent is N2.
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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