\(BaC{l_2}.x{H_2}O + {H_2}S{O_4}\)\( \to \)\(BaS{O_{4 \downarrow }} + 2HCl + x{H_2}O\)
1 mol 1 mol
\(\dfrac{{1,952}}{M} = {8.10^{ - 3}}\) (mol) \(\dfrac{{1,864}}{{233}} = {8.10^{ - 3}}\) mol
\( \Rightarrow M = 244g/mol = {M_{BaC{l_2}.x{H_2}O}}\). Từ đó :
\( \Rightarrow x = \dfrac{{244 - 208}}{{18}} = 2\).
Đáp số : BaCl2.2H2O
