CTHH: M(OH)2
Giả sử số mol bazo là 1 (mol)
\(=>m_{M\left(OH\right)_2}=M+34\left(g\right)\)
PTHH: \(M\left(OH\right)_2+H_2SO_4\rightarrow MSO_4+2H_2O\)
_________1------------>1------------>1_______________(mol)
=> \(m_{MSO_4}=M_M+96\left(g\right)\)
=> \(m_{H_2SO_4}=1.98=98\left(g\right)\)
=> m dd H2SO4 = \(\frac{98.100}{20}=490\left(g\right)\)
=> \(C\%=\frac{M+96}{490+M+34}.100\%=21,9\%\)
=> MM = 24 (g/mol) => M là Mg
=> CTHH: Mg(OH)2