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]

$\begin{matrix} (\text{NH}$4)2\text{Cr}2\text{O}7 & \xrightarrow{\Delta} & \text{N}2 & + & 4\text{H}2\text{O} & + & \text{Cr}2\text{O}3 \ 2[14 + 4(1)] & & & & & & 2(52) \ + 2(52) + 7(16) & & & & & & + 3(16) \ = 36 + 104 & & & & & & = 104 + 48 \ + 112 = 252 \text{ g} & & & & & & = 152 \text{ g} \ \end{matrix}

(i) 252 g of (NH₄)₂Cr₂O₇ = 1 mole

∴ 63 g of (NH₄)₂Cr₂O₇ = $\dfrac{1}{252}$ x 63 = 0.25 moles

Hence, no. of moles = 0.25 moles

(ii)

$\begin{matrix} (\text{NH}$4)2\text{Cr}2\text{O}7 & : & \text{N}_2 \ 1 \text{ mol.} & : & 1 \text{ mol.} \ 0.25 \text{ mol.} & : & x \ \end{matrix}

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 x 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 x 0.25 = 38 g.

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