a,
$m_{dd\text{ban đầu}}=100.1,2=120g$
$\Rightarrow n_{NaOH\text{bđ}}=\dfrac{120.20\%}{40}=0,6(mol)$
Sau phản ứng, $m_{dd}=120+4,6=124,6g$
$\Rightarrow n_{NaOH\text{dư}}=\dfrac{124,6.14,1255\%}{40}=0,44(mol)$
$\Rightarrow n_{NaOH\text{pứ}}=0,6-0,44=0,16(mol)$
Dư $NaOH$ nên tạo muối $Na_2CO_3$.
$2NaOH+CO_2\to Na_2CO_3+H_2O$
$\Rightarrow n_C=n_{CO_2}=0,08(mol)$
Ta có: $m_{CO_2}+m_{H_2O}=4,6g$
$\Rightarrow n_{H_2O}=\dfrac{4,6-0,08.44}{18}=0,06(mol)$
$\Rightarrow n_H=2n_{H_2O}=0,12(mol)$
$n_O=\dfrac{3-0,12-0,08.12}{16}=0,12(mol)$
$n_C : n_H: n_O=0,08: 0,12:0,12=2:3:3$
$\Rightarrow$ CTĐGN $C_2H_3O_3$
b,
Đặt CTPT: $(C_2H_3O_3)_n$
$\Rightarrow 100<(12.2+3+16.3)n <200$
$\Leftrightarrow 1,33<n<2,67$
$\Rightarrow n=2$
Vậy CTPT là $C_4H_6O_6$