Question

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.

Answer 1

Given, m_N2 =28g;

m_H2 = 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