\(n_M=\frac{0,8}{M_M}\left(mol\right)\)
\(n_{H_2SO_4}=0,5.0,1=0,05\left(mol\right)\)
\(n_{NaOH}=0,0334.1=0,0334\left(mol\right)\)
PTHH: \(M+H_2SO_4\rightarrow MSO_4+H_2\)
______\(\frac{0,8}{M_M}\)--->\(\frac{0,8}{M_M}\)_____________________(mol)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,0334----->0,0167_______________________(mol)
=> \(\frac{0,8}{M_M}+0,0167=0,05\) => MM = 24(g/mol)
=> M là Mg