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JEE Main 2019 · Chemistry · s-Block, p-Block Elements and Hydrogen · medium

Phosphorous acid, H3PO3, on heating disproportionates to H3PO4 and PH3. Starting with 8.2 g of H3PO3 (molar mass 82) and complete reaction, the mass…

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Phosphorous acid, H3PO3\mathrm{H_3PO_3}, on heating disproportionates to H3PO4\mathrm{H_3PO_4} and PH3\mathrm{PH_3}. Starting with 8.2 g8.2\ \mathrm{g} of H3PO3\mathrm{H_3PO_3} (molar mass 8282) and complete reaction, the mass of H3PO4\mathrm{H_3PO_4} (molar mass 9898) formed is
  1. A. 9.80 g9.80\ \mathrm{g}
  2. B. 2.45 g2.45\ \mathrm{g}
  3. C. 7.35 g7.35\ \mathrm{g}
  4. D. 4.90 g4.90\ \mathrm{g}
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Answer: (C) 7.35 g7.35\ \mathrm{g}

Balanced equation: 4H3PO3→Δ3H3PO4+PH34\mathrm{H_3PO_3} \xrightarrow{\Delta} 3\mathrm{H_3PO_4} + \mathrm{PH_3} (P goes from +3+3 to +5+5 in three atoms and to −3-3 in one: 3×2=63\times2 = 6 electrons lost, 66 gained). Moles of H3PO3=8.2/82=0.100\mathrm{H_3PO_3} = 8.2/82 = 0.100. Moles of H3PO4=34×0.100=0.075\mathrm{H_3PO_4} = \tfrac34\times0.100 = 0.075, mass =0.075×98=7.35 g=0.075\times98 = 7.35\ \mathrm{g}. Taking a 1:1 ratio would give 9.80 g9.80\ \mathrm{g}.
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