From the equation :
(NH₄)₂Cr₂O₇ ⟶ N₂(g) + 4H₂O(g) + Cr₂O₃
Calculate:
(i) the quantity in moles of (NH₄)₂Cr₂O₇ if 63 gm of
(NH₄)₂Cr₂O₇ is heated.
(ii) the quantity in moles of N₂ formed.
(iii) the volume in litres or dm3 of N₂ evolved at s.t.p.
(iv) the mass in grams of Cr₂O₃ formed at the same time.
[H = 1, Cr = 52, N = 14]

(NH₄)₂Cr₂O₇ ⟶ N₂(g) + 4H₂O(g) + Cr₂O₃

252 g                                                 152 g

(i) 252 g of (NH₄)₂Cr₂O₇ = 1 mole

∴ 63 g of (NH₄)₂Cr₂O₇ = 1\over 252 × 63

= 0.25 moles

Hence, no. of moles = 0.25 moles

(ii) 1 mole of (NH₄)₂Cr₂O₇ produces  = 1 mole on N₂

Hence, 0.25 moles of (NH₄)₂Cr₂O₇ will produce 0.25 moles of nitrogen.

(iii) 1 mole of N₂ occupies 22.4 lit.

∴ 0.25 moles of N₂ will occupy = 22.4 × 0.25 = 5.6 lit.

Hence, volume of N₂ evolved at s.t.p = 5.6 lit.

(iv) 1 mole of Cr₂O₇ = 152 g.

∴ 0.25 moles of Cr₂O₇ = 152 × 0.25 = 38 g.

Hence, mass in gms of Cr₂O₇ formed = 38 g.