ICSE Class 10 Answered

(a)  The given reaction is as follows:

(i)    Given:

       Weight of (NH₄)₂Cr₂O₇ = 63 gm

        Molar mass of (NH₄)₂Cr₂O₇

         = (2 ×14) + (8 × 1) + (2 × 52) + (7 × 16)

         = 28 + 8 + 104 + 112

         = 252 gm

    1 mole (NH₄)₂Cr₂O₇ = 252 gm

    Hence, 63 gm of (NH₄)₂Cr₂O₇ = 0.25 moles

    The quantity of moles of (NH₄)₂Cr₂O₇ if 63 gm of (NH₄)₂Cr₂O₇ is heated is 0.25 moles.

(ii)  From the given chemical equation, 1 mole of (NH₄)₂Cr₂O₇ produces 1 mole of nitrogen gas.

      Hence, 0.25 moles of (NH₄)₂Cr₂O₇ can produce 0.25 moles of nitrogen gas.

       The quantity in moles of nitrogen formed is 0.25 moles.

(iii) One mole of an ideal gas at S.T.P. occupies 22.4 litres or dm³.

      Hence, 0.25 moles of (NH₄)₂Cr₂O₇ will occupy 0.25 ´ 22.4 = 5.6 litres or dm³.

      The volume in litres or dm³ of N₂ evolved at S.T.P. is 5.6 litres or dm³.

(iv) From the given chemical equation, 1 mole of (NH₄)₂Cr₂O₇ produces 1 mole of Cr₂O₃.

       Hence, 0.25 moles of (NH₄)₂Cr₂O₇ will produce 0.25 moles of Cr₂O₃.

      Molar mass of Cr₂O₃ = (2 × 52) + (3 × 16)

                                    = 104 + 48

                                     = 152 gm

 (v) 1 mole Cr₂O₃ = 152 gm

       Hence, 0.25 moles of Cr₂O₃ = 0.25 × 152 = 38 gm

       The mass in grams of Cr₂O₃ formed at the same time is 38 gm.