4 B(s)+3O2(g)⟶ΔI2Al(s)+N2(g)⟶ΔIIHere, I and II are

4 B(s)+3O2(g)ΔI

2Al(s)+N2(g)ΔII

Here, I and II are

  1. A

    IB2O3(s),IIAlN(g)

  2. B

    IB2O3(g),IIAlN(s)

  3. C

    IB2O3( g),IIAlN(g)

  4. D

    IB2O3(s),IIAlN(s)

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    Solution:

    4 B(s)+3O2(g)Δ2 B2O3(s)

    2Al(s)+N2(g)Δ2AlN(s)

     

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