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JEE Main 2018 · Chemistry · s-Block, p-Block Elements and Hydrogen · easy

In the Solvay process, 2NaCl→ Na2CO3 overall. Starting from 1.00 kg of pure NaCl (M=58.5) and an overall yield of 80%, the mass of anhydrous sodium…

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In the Solvay process, 2NaCl→Na2CO3\mathrm{2NaCl\to Na_2CO_3} overall. Starting from 1.00 kg1.00\ \mathrm{kg} of pure NaCl\mathrm{NaCl} (M=58.5M=58.5) and an overall yield of 80%80\%, the mass of anhydrous sodium carbonate (M=106M=106) obtained is about
  1. A. 453 g453\ \mathrm{g}
  2. B. 725 g725\ \mathrm{g}
  3. C. 906 g906\ \mathrm{g}
  4. D. 1449 g1449\ \mathrm{g}
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Answer: (B) 725 g725\ \mathrm{g}

Moles of NaCl =1000/58.5=17.09=1000/58.5=17.09. Two NaCl give one Na2CO3\mathrm{Na_2CO_3}, so the theoretical yield is 8.5478.547 mol =8.547×106=906=8.547\times106=906 g. At 80%80\% yield: 0.80×906=7250.80\times906=725 g. Forgetting the factor 12\tfrac12 gives 14491449 g and omitting the yield gives 906906 g.
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