Match the amount of reactant given in column I with neutralisation reactions given in column II.
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Solution
(A) 1. mEq of NaOH =200×0.5=100mEq≡100mEq of H2SO4 2. 4.9gH2+6SO4=4.998×103 =50mmol≡100mEq (n factor =2) ≡200mmol of O atoms. (1molH2SO4≡4molOatoms) 3.. Highest oxidation state of S=+6 4. Hybridisation=sp3; geometry and shape ⇒Tetrahedral
(B) 4.9gH3PO4≡4.998×1000=50mmol =50×3 (n factor)≡150mEq 2.50mmol of H3PO4≡50×4=200mmol of O atoms). (1molH3PO4≡4mol O atoms) 1. But 150 mEq H3PO4≠mEq of NaOH 5. Hybridisation is sp3, so geometry and shape is tetrahedral. 3. Maximum oxidation state of P=+5
(C) 1.4.9gH2C2O4≡4.990×1000 =50mmol=100mEq (n factor =2) 2. 50mmoles of H2C2O4≡50×4=200mmolesO atoms (1molH2C2O4=4mol o atoms) ≡100mEqNaOH 3. Highest oxidation state of C=+4 4. Reacts with oxidising agents such as MNO⊖4 and Cr2O2−7.
(D) 1.5.3g of Na2CO)3=5.3106×1000 =50mmoles 2 Na2CO3 does not react with NaOH. 3. Highest oxidation states of C=+4 5.CO2−3, hybridisation is sp2, so geometry and shape is planar. 7. 1molNa2CO3≡3mol O atoms.