Tamil Nadu Board of Secondary EducationHSC Science Class 11th

# 28 g of Nitrogen and 6 g of hydrogen were mixed In a 1 litre closed container. At equIlibrium 17 g NH3 was produced. Calculate the weight of nitrogen, hydrogen at equilibrium. - Chemistry

Numerical

28 g of Nitrogen and 6 g of hydrogen were mixed In a 1 litre closed container. At equIlibrium 17 g NH3 was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.

#### Solution

Given, mN2 =28g;

mH2 = 6g;

V = 1 L.

("n"_("N"_2))_"initial" = 28/28 = 1

("n"_("H"_2))_"initial" = 6/2 = 3

$\ce{N2(g) + 3 H2(g) <=> 2NH3(g)}$

 N2 (g) H2 (g) NH3 (g) Initial concentrations 1 3 - Reacted 0.5 1.5 - Equilibrium concentration 0.5 1.5 1

[NH3] = 17/17 = 1 mol

Weight of N2 = (no. of moles of N2) × (molar mass of N2)

= 0.5 × 28 = 14g

Weight of H2 = (no. of moles of H2) × (molar mass of H2)

= 1.5 × 2 = 3g

Concept: Physical and Chemical Equilibrium
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#### APPEARS IN

Tamil Nadu Board Samacheer Kalvi Class 11th Chemistry Volume 1 and 2 Answers Guide
Chapter 8 Physical and Chemical Equilibrium
Evaluation | Q II. 20. | Page 27
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