CTHH: BO
Gọi số mol BO là a(mol)
PTHH: BO + H2SO4 --> BSO4 + H2O
a--->a----------->a
=> \(m_{H_2SO_4}=98a\left(g\right)\)
=> \(m_{ddH_2SO_4}=\dfrac{98a.100}{10}=980a\left(g\right)\)
mdd sau pư = a(MB + 16) + 980a (g)
\(m_{BSO_4}=a\left(M_B+96\right)\left(g\right)\)
=> \(C\%=\dfrac{a\left(M_B+96\right)}{a\left(M_B+16\right)+980a}.100\%=11,765\%\)
=> MB = 24 (g/mol)
=> B là Mg