1.24g white phosphorous P₄ was allowed to react with O₂ to produce P₄O₆ an acidic oxide.The solution of P₄O₆​ in water is acidic and is neutralized with NaOH.How many moles of NaOH are required to neutralize H₃ PO₃ generated in the above case?

The chemical reactions mentioned in above case are :

 P4 + 3O2 → P4O6

P4O6 + 6H2O → 4H3PO3

2NaOH + H3PO3→ Na2HPO3 + 2H2O

Moles of NaOH (nNaOH) can be calculated as :

1 mol P4 ≡ 1 mol P4O6 ≡ 4 mol H3PO3

1 mol H3PO3 ≡ 2 mol Na OH

nNaOH  =

  = 1.24 g P4 * 1mol P4/124g P4 * 1mol of P4O6/1 mol P4

   * 2 mol NaOH/1 mol of H3PO3

  = 0.01 × 4 × 2

  = 0.08 mol