Question

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

Answer 1

Answer:

NH_{​​​​​3} : 0.380 mol (produced)

H_{​​​​​​2} : 0.034 mol (remains)

N_{​​​​​​2} : 0 mol (remains)

Nitrogen is the limiting reagent.

Explanation:

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

In the whole process 0.380 mol of ammonia produced.

All the 0.190 mol of N_{​​​​​​2} is consumed and 0 mol of N_{​​​​​​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 of H_{​​​​​​2} remains.

Now the limiting reagent,

Limiting reagent is the reagent in the reaction which gets completely consumed in the process. In this reaction N_{​​​​​​2} is the reactant that hets completely consumed after the complete consumption of N_{​​​​​2} no further production of ammonia can take place. Thereby, the amount of N_{​​​​​​2} limits the production of ammonia.

So, the limiting reagent is N_{​​​​​​2.}