= Solution
Protonolysis of both magnesium–butyl bonds gives
$$
\mathrm{Mg(nBu)_2+2HN(SiMe_3)_2\longrightarrow Mg[N(SiMe_3)_2]_2+2nBuH}.
$$
Thus \b[A is magnesium bis(hexamethyldisilazide), and the gas is butane]. Addition of the corresponding lithium amide gives \b[B = $\mathrm{LiMg[N(SiMe_3)_2]_3}$], a lithium tris(amido)magnesiate.
A useful connectivity sketch is $\mathrm{(R_2N)Mg(\mu-NR_2)_2Li}$ with $R=\mathrm{SiMe_3}$: magnesium binds three approximately coplanar nitrogen atoms, two of which also bridge to lithium. The remaining nitrogen is terminal at magnesium. Lithium can gain additional intramolecular contacts to methyl groups on the silylamide substituents. These contacts and the steric envelope of the bulky ligands help satisfy coordination without intermolecular aggregation. “Monomeric” here means one LiMg formula unit, not a single metal atom or an isolated bare $\mathrm{Li^+}$ counterion.
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