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JEE Main 2020 · Chemistry · s-Block, p-Block Elements and Hydrogen · easy

On heating, phosphorous acid disproportionates as 4H3PO3→3H3PO4+PH3. If 8.2 g of H3PO3 (molar mass 82 g mol^-1) is heated completely, the mass of PH3…

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On heating, phosphorous acid disproportionates as 4H3PO3→3H3PO4+PH34\mathrm{H_3PO_3}\rightarrow3\mathrm{H_3PO_4}+\mathrm{PH_3}. If 8.2 g8.2\ \mathrm{g} of H3PO3\mathrm{H_3PO_3} (molar mass 82 g mol−182\ \mathrm{g\,mol^{-1}}) is heated completely, the mass of PH3\mathrm{PH_3} (34 g mol−134\ \mathrm{g\,mol^{-1}}) formed is
  1. A. 0.85 g0.85\ \mathrm{g}
  2. B. 3.40 g3.40\ \mathrm{g}
  3. C. 0.34 g0.34\ \mathrm{g}
  4. D. 2.55 g2.55\ \mathrm{g}
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Answer: (A) 0.85 g0.85\ \mathrm{g}

Moles of H3PO3=8.2/82=0.10\mathrm{H_3PO_3}=8.2/82=0.10. From the equation 44 mol give 11 mol PH3\mathrm{PH_3}, so PH3=0.025\mathrm{PH_3}=0.025 mol, mass =0.025×34=0.85=0.025\times34=0.85 g. (Oxidation states: P goes +3→+5+3\to+5 in 3 molecules and +3→−3+3\to-3 in 1; electrons balance: 3×2=63\times2=6.)
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