Bình 1: m_{H_2O}=5,4(g)
Bảo toàn H: n_H=2n_{H_2O}=2.{5,4}/{18}=0,6(mol)
Bình 2: m_{CaCO_3}=20(g)
Bảo toàn C: n_C=n_{CO_2}=n_{CaCO_3}={20}/{100}=0,2(mol)
->n_O={4,6-0,6-0,2.12}/{16}=0,1(mol)
->n_C:n_H:n_O=0,2:0,6:0,1=2:6:1
-> CTTN: (C_2H_6O)_n
Mà M_A=23.2=46
->(12.2+6+16).n=46
->n=1
->A:\ C_2H_6O
