Solid ammonium dichromate decomposes as under:

(NH₄)₂Cr₂O₇ ⟶ N₂ + Cr₂O₃ + 4H₂O

If 63 g of ammonium dichromate decomposes. Calculate

(a) the quantity in moles of (NH₄)₂Cr₂O₇

(b) the quantity in moles of nitrogen formed.

(c) the volume of N₂ evolved at S.T.P.

(d) the loss of mass

(iv) the mass of chromium (III) formed at the same time.

$\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}

(a) 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

(b)

$\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.

(c) 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.

(d) 252 g of (NH₄)₂Cr₂O₇ decomposes to give 152 g of solid Cr₂O₃, loss in mass = 252 - 152 = 100 g

If 63 g of (NH₄)₂Cr₂O₇ decomposes then loss in mass is

$\dfrac{100}{252}$ x 63 = 25 g

(e) 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.