Bạn tham khảo:
TH1: Tạo $Na_2SO_3$
$2NaOH+SO_2 \to Na_2SO_3+H_2O$
$n_{Na_2SO_3}=0,022(mol)$
$n_{NaOH}=0,03(mol)$
$\to n_{Na_2SO_3}=0,15(mol)$
$\to$ TH1 loại
TH2: Tạo $NaHSO_3$
$NaOH+SO_2 \to NaHSO_3$
$n_{NaHSO_3}=0,026(mol)$
$n_{NaHSO_3}=0,03(mol)$
$\to$ TH2 loại
TH3: Tạo 2 muối
$2NaOH+SO_2 \to Na_2SO_3+H_2O(1)$
$NaOH+SO_2 \to NaHSO_3(2)$
$n_{SO_2(1)}=a(mol)$
$n_{SO_2(2)}=b(mol)$
$n_{NaOH}=2a+b=0,03(1)$
$m_{khan}=126a+104b=2,71(2)$
$a=0,005$
$b=0,02$
$2R+2nH_2SO_4 \xrightarrow{t^{o}} R_2(SO_4)_n+nSO_2+2nH_2O$
$∑n_{SO_2}=0,005+0,02=0,025(mol)$
$\to n_R=\frac{0,025.2}{n}=\frac{0,05}{n}(mol)$
$M_R=\frac{1,6.n}{0,05}=32.n(g/mol)$
$n=2; R=64$