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ICSE Class 10 Chemistry Question 17 of 28

Stoichiometry — Percentage Composition — Empirical & Molecular Formula — Chemical Equations — Question 25

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25
Question

Question 1(2015)

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]

Answer

(NH4)2Cr2O7→ΔN2+4H2O+Cr2O32[14+4(1)]2(52)+2(52)+7(16)+3(16)=36+104=104+48+112=252 g=152 g\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 = 1252\dfrac{1}{252} x 63 = 0.25 moles

Hence, no. of moles = 0.25 moles

(ii)

(NH4)2Cr2O7:N21 mol.:1 mol.0.25 mol.:x\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.