Giả sử có 1 mol M2CO3.10H2O \(\Rightarrow\) nM2CO3 = 1mol
M2CO3 + BaCl2 \(\rightarrow\) BaCO3 + 2MCl (1)
1mol 1mol 1 mol 2mol
\(\Rightarrow\) m ddBaCl2 = 1.208. \(\frac{100}{5}\)= 4160 (g)
\(\Rightarrow\) \(\frac{2.\left(M+35,5\right)}{4160+1.\left(2M+240\right)-1.197}\) = \(\frac{2,7536}{100}\)
\(\Rightarrow\) M = 23 \(\Rightarrow\) M là Na
\(\Rightarrow\) M2CO3.10H2O