For a reaction, $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})$; identify dihydrogen $\left(\mathrm{H}_{2}\right)$ as a limiting reagent in the following reaction mixtures.

(C) 56 g of $\mathrm{N}_{2}+10 \mathrm{~g}$ of $\mathrm{H}_{2}$

(A) 14 g of $\mathrm{N}_{2}+4 \mathrm{~g}$ of $\mathrm{H}_{2}$

(B) 28 g of $\mathrm{N}_{2}+6 \mathrm{~g}$ of $\mathrm{H}_{2}$

(D) 35 g of $\mathrm{N}_{2}+8 \mathrm{~g}$ of $\mathrm{H}_{2}$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Some Basic Concepts of Chemistry

For a reaction,

N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g)

; identify dihydrogen

(H_{2})

as a limiting reagent in the following reaction mixtures.

(A)14 g of $N_{2} + 4 g$ of $H_{2}$
(B)28 g of $N_{2} + 6 g$ of $H_{2}$
(C)56 g of $N_{2} + 10 g$ of $H_{2}$ ✓ Correct
(D)35 g of $N_{2} + 8 g$ of $H_{2}$

Answer: (C)

Solution

For

N X_{2} + 3 H X_{2} 2 NH X_{3}

every mole of nitrogen needs three of hydrogen. Taking

56

g of

N X_{2}

gives

\frac{56}{28} = 2

mol and

10

g of

H X_{2}

gives

\frac{10}{2} = 5

mol. Two moles of nitrogen would call for

6

mol of hydrogen, but only

5

mol are present, so hydrogen runs out first and is the limiting reagent. In the other mixtures hydrogen is in excess or nitrogen is the one that limits.

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Some Basic Concepts of Chemistry

Solution

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