Chemistry Labs

Problem 2

Resistive gas sensors using semiconducting metal oxides (SMOX) detect tiny amounts of impurities. The mixed oxide X, which has the normal spinel structure A2+B23+O42−\mathrm{A^{2+}B^{3+}_2O^{2-}_4}, is obtained by decomposing in air the crystalline hydrated metal oxalate ZCX2OX4 ⋅ kHX2O\ce{ZC2O4*\mathit{k}H2O} of metal Z. When heated to 140 °C the hydrate loses 19.7 % of its mass; further heating in air at 500 °C yields 2.407 g of black X together with 3.8 dm3^3 of COX2\ce{CO2} (at 101325 Pa, 500 °C). (a) Determine the formula of X and the value of kk. (b) In the spinel structure the OX2−\ce{O^{2-}} ions form a face-centred cubic lattice; A2+\mathrm{A^{2+}} cations occupy part of the tetrahedral voids and B3+\mathrm{B^{3+}} cations part of the octahedral voids. Calculate the percentage of tetrahedral sites occupied in a normal spinel AB2O4\mathrm{AB_2O_4}.
Step 1 of 4: Moles of CO2 released
n(COX2)=pVRT=101325×3.8×10−38.314×773=0.060 moln(\ce{CO2}) = \dfrac{pV}{RT} = \dfrac{101325\times 3.8\times 10^{-3}}{8.314\times 773} = 0.060\ \text{mol}
Analysis

The gas is collected hot, so the ideal gas law at T=500T = 500 °C =773= 773 K gives n(COX2)=0.060n(\ce{CO2}) = 0.060 mol, corresponding to 0.060/6=0.0100.060/6 = 0.010 mol of ZX3OX4\ce{Z3O4} formed.

Common pitfall. Use T = 773 K (the 500 °C decomposition temperature), not 298 K, when applying pV = nRT.