A: $R(HCO_3)_n$
$n_A=\dfrac{9,875}{R+61n}$
$\Rightarrow n_{R_2(SO_4)_n}=0,5n_A=\dfrac{4,9375}{R+61n}$
$\Rightarrow \dfrac{4,9375}{R+61n}=\dfrac{8,25}{2R+96n}$
$\Leftrightarrow 8,25(R+61n)=4,9375(2R+96n)$
$\Leftrightarrow R=18n$
$\Rightarrow n=1; R=18(NH_4)$
Vậy A là $NH_4HCO_3$
$NH_4HCO_3+HNO_3\to NH_4NO_3+CO_2+H_2O$
$n_{NH_4NO_3}=n_A=0,125(mol)$
$\Rightarrow m_{NH_4NO_3}=0,125.80=10g<m_B$
$\Rightarrow$ B là muối bị hidrat hoá.
$n_{H_2O}=\dfrac{23,5-10}{18}=0,75(mol)$
$n_{NH_4NO_3}:n_{H_2O}=0,125:0,75=1:6$
$\Rightarrow B$ là $NH_4NO_3.6H_2O$
