$a)\\ E:C_xH_yO_z\\ m_{H_2SO_4}=100.96,48\%=96,48(g)\\ C\%_{H_2SO_4\ 90\%}=\dfrac{96,48}{100+m_{H_2O}}.100\%=90\%\\ \to m_{H_2O}=7,2(g)\\ \xrightarrow{btnt\ H:} n_H=2n_{H_2O}=\dfrac{2.7,2}{18}=0,8(mol)\\ \xrightarrow{btnt\ C:} n_C=n_{CO_2}=n_{CaCO_3}=\dfrac{40}{100}=0,4(mol)\\ \xrightarrow{btnt\ O:} n_{O(E)}+2n_{O_2}=2n_{CO_2}+n_{H_2O}\\ \to n_{O(E)}+\dfrac{2.8,96}{22,4}=2.0,4+\dfrac{0,8}{2}\\ \to n_{O(E)}=0,4(mol)\\ x=\dfrac{0,4}{0,2}=2\\ y=\dfrac{0,8}{0,2}=4\\ z=\dfrac{0,4}{0,2}=2\\ E: C_2H_4O_2\\ b)\\ n_{NaOH}=\dfrac{12,8}{40}=0,32(mol)$

· Trường hợp 1: E là $\rm CH_3COOH$

$CH_3COOH+NaOH\to CH_3COONa+H_2O\\ n_{CH_3COOH}<n_{NaOH}\to NaOH\ du\ spu\\ n_{CH_3COONa}=n_{NaOH(pu)}=n_{CH_3COOH}=0,2(mol)\\ m=12,8-0,2.40+0,2.82=21,2(g)$

· Trường hợp 2: E là $\rm HCOOCH_3$

$HCOOCH_3+NaOH\xrightarrow{t^o} HOONa+H_2O\\ n_{CH_3COOH}<n_{NaOH}\to NaOH\ du\ spu\\ n_{HCOONa}=n_{NaOH(pu)}=n_{CH_3COOH}=0,2(mol)\\ m=12,8-0,2.40+0,2.68=18,4(g)$