${\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7 .10{\mathrm{H}}_2\mathrm{O} \stackrel{\mathrm{conc}. \mathrm{HCl}}{\to } \mathrm{A}.$

Compound 'A' is

The reaction is given below:

${\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7 .10{\mathrm{H}}_2\mathrm{O} \stackrel{\mathrm{conc}. \mathrm{HCl}}{\to } 4{\mathrm{H}}_3{\mathrm{BO}}_3 + 2\mathrm{NaCl} + 5{\mathrm{H}}_2\mathrm{O}.$

So, the compound A is ${\mathrm{H}}_3{\mathrm{BO}}_3$.