A black powder (A) when heated with $\mathrm{NaCl}$ and conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ gives off a greenish yellow gas (B). Gas 'B' on passing through boiling $\mathrm{KOH}$ yields compounds, one of which when heated with the black powder evolves oxygen. (A) and (B) respectively are

Color of MnO₂ is black-brown and color of Pb₃O₄ is red. Therefore, A is MnO₂. Reaction of MnO₂ with NaCl and concentrated H₂SO₄ is:

${\mathrm{MnO}}_2+2\mathrm{NaCl}+3{\mathrm{H}}_2{\mathrm{SO}}_4\to 2{\mathrm{NaHSO}}_4+{\mathrm{MnSO}}_4+\underset{\mathrm{green} \mathrm{color} \mathrm{gas}}{{\mathrm{Cl}}_2}\uparrow +2{\mathrm{H}}_2\mathrm{O}$

Passing of chlorine gas with KOH:

$3{\mathrm{Cl}}_2+6\mathrm{KOH}\to {\mathrm{KClO}}_3+5\mathrm{KCl}+3{\mathrm{H}}_2\mathrm{O}$

When KClO₃ is treated with black powder (or MnO₂), liberates oxygen gas.

$2{\mathrm{KClO}}_3\to 2\mathrm{KCl}+3{\mathrm{O}}_2\uparrow$