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Some Basic Concepts of Chemistry question

2022 · 29 Jul · Shift 1 · Q2

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Some Basic Concepts of Chemistry
2022 · 29 Jul · Shift 1 · Q2

Some Basic Concepts of Chemistry question

2022 · 29 Jul · Shift 1 · Q2

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1

\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}

20 \mathrm{~g} \quad ~~~5 \mathrm{~g}

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

\mathrm{NH}_{3}

formed respectively are :

A
$\mathrm{H}_{2}, 1.42$ moles
B
$\mathrm{H}_{2}, 0.71$ moles
C
$\mathrm{N}_{2}, 1.42$ moles
D
$\mathrm{N}_{2}, 0.71$ moles

View written solutionFree

Correct answer: C

Write the balanced reaction

$\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightarrow 2\mathrm{NH_3(g)}$

Given masses:

$\mathrm{N_2} = 20\,\text{g}$
$\mathrm{H_2} = 5\,\text{g}$

Calculate moles of reactants

For nitrogen: $n(\mathrm{N_2}) = \frac{20}{28} = 0.714\,\text{mol}$

For hydrogen: $n(\mathrm{H_2}) = \frac{5}{2} = 2.5\,\text{mol}$

Find the limiting reagent

From the balanced equation: $1\,\text{mol } \mathrm{N_2} \text{ requires } 3\,\text{mol } \mathrm{H_2}$

So, $0.714$ mol $\mathrm{N_2}$ requires: $0.714 \times 3 = 2.142\,\text{mol } \mathrm{H_2}$

Available $\mathrm{H_2} = 2.5$ mol, which is more than required.

Therefore, $\mathrm{N_2}$ is the limiting reagent.

Calculate moles of ammonia formed

From the reaction: $1\,\text{mol } \mathrm{N_2} \rightarrow 2\,\text{mol } \mathrm{NH_3}$

Thus, $0.714\,\text{mol } \mathrm{N_2} \rightarrow 0.714 \times 2 = 1.428\,\text{mol } \mathrm{NH_3}$

So, moles of ammonia formed $\approx 1.42$ mol.

Match with options

Limiting reagent = $\mathrm{N_2}$
Moles of $\mathrm{NH_3}$ formed = $1.42$

Hence, the correct option is:

$\boxed{\text{C}}$

Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.

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