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Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2024 — 4 April, Shift 1 — Question 76

Xg\mathrm{X} g of ethylamine is subjected to reaction with NaNO2/HCl\mathrm{NaNO}_{2} / \mathrm{HCl} followed by water and evolved dinitrogen gas which occupied 2.24 L2.24\ \mathrm{L} volume at STP. The value of XX is _____\_\_\_\_\_ ×10−1 g\times 10^{-1} \mathrm{~g}.

Answer: 45

Numerical answer — enter this value.

Step-by-step solution

Primary amine reaction: C2H5NH2+HNO2→C2H5OH+N2+H2O\mathrm{C_2H_5NH_2 + HNO_2 \rightarrow C_2H_5OH + N_2 + H_2O}

1 mol amine → 1 mol N2\mathrm{N_2}

At STP: 22.4 L=1 mol\mathrm{22.4\ L = 1\ mol}

Thus moles of N2\mathrm{N_2} =2.2422.4=0.1 mol= \mathrm{\frac{2.24}{22.4} = 0.1\ mol}

So moles of ethylamine =0.1 mol= \mathrm{0.1\ mol}

Molar mass of ethylamine =45 g mol−1= \mathrm{45\ g\ mol^{-1}}

Mass =0.1×45=4.5 g= \mathrm{0.1 \times 45 = 4.5\ g}

Expressed as, 45×10−1 g\mathrm{45 \times 10^{-1}\ g}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
4 April, Shift 1
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic
Other Miscellaneous Reactions and Related Problems

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