JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Some Basic Concepts of Chemistry

N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 NH_{3 (g)}

20 g 5 g

Consider the above reaction, the limiting reagent of the reaction and number of moles of

NH_{3}

formed respectively are:

Answer: (C)

Solution

Calculate moles:

n (N_{2}) = 20/28 = 0.714

mol,

n (H_{2}) = 5/2 = 2.5

mol. For stoichiometry, divide by coefficients:

N_{2} : 0.714/1 = 0.714

;

H_{2} : 2.5/3 = 0.833

. Since 0.714 < 0.833,

N_{2}

is the limiting reagent. Moles of

NH_{3}

formed =

2 \times 0.714 = 1.428 \approx 1.42

moles.

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option