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JEE Main 2022 · Chemistry · Mole Concept and Stoichiometry · medium

A 4.00 g mixture of NaHCO3 and Na2CO3 loses 0.62 g in mass on strong heating. The residue is then completely neutralised with 0.50 M HCl (all…

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A 4.00 g4.00\ \mathrm{g} mixture of NaHCO3\mathrm{NaHCO_3} and Na2CO3\mathrm{Na_2CO_3} loses 0.62 g0.62\ \mathrm{g} in mass on strong heating. The residue is then completely neutralised with 0.50 M0.50\ \mathrm{M} HCl (all carbonate converted to CO2\mathrm{CO_2}). Volume of HCl consumed is (molar masses in g mol−1\mathrm{g\,mol^{-1}}: NaHCO3=84\mathrm{NaHCO_3}=84, Na2CO3=106\mathrm{Na_2CO_3}=106)
  1. A. 127.5 mL127.5\ \mathrm{mL}
  2. B. 63.8 mL63.8\ \mathrm{mL}
  3. C. 95.2 mL95.2\ \mathrm{mL}
  4. D. 167.5 mL167.5\ \mathrm{mL}
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Answer: (A) 127.5 mL127.5\ \mathrm{mL}

Only 2NaHCO3→Na2CO3+CO2+H2O2\mathrm{NaHCO_3}\to \mathrm{Na_2CO_3}+\mathrm{CO_2}+\mathrm{H_2O} causes mass loss; 62 g62\ \mathrm g is lost per 168 g168\ \mathrm g of bicarbonate. So NaHCO3=0.62×168/62=1.68 g\mathrm{NaHCO_3}=0.62\times168/62=1.68\ \mathrm g (0.020.02 mol) and original Na2CO3=2.32 g\mathrm{Na_2CO_3}=2.32\ \mathrm g (0.021890.02189 mol). The residue carbonate is 0.02189+0.01=0.031890.02189+0.01=0.03189 mol, needing 2×0.03189=0.063772\times0.03189=0.06377 mol HCl. Shortcut: sodium is conserved, so HCl needed equals 0.02+2(0.02189)=0.06380.02+2(0.02189)=0.0638 mol. Volume =0.0638/0.50=127.5 mL=0.0638/0.50=127.5\ \mathrm{mL}.
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