Em tham khảo nha:
\(\begin{array}{l}
5)\\
PTTQ:\\
R + {H_2}S{O_4} \to RS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{0,896}}{{22,4}} = 0,04\,mol\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,04\,mol\\
BTKL:{m_{hh}} + {m_{{H_2}S{O_4}}} = {m_m} + {m_{{H_2}}}\\
\Rightarrow {m_m} = 1,78 + 0,04 \times 98 - 0,04 \times 2 = 5,62g\\
6)\\
T{N_1}:\\
M + 2HCl \to MC{l_2} + {H_2}\\
{n_{HCl}} = 0,1 \times 0,75 = 0,075\,mol\\
\text{ M dư sau phản ứng } \Rightarrow {n_M} > \dfrac{{0,075}}{2} \Rightarrow {n_M} > 0,0375\,mol\\
\Rightarrow {M_M} < \dfrac{{1,2}}{{0,0375}} \Rightarrow {M_M} < 32\,g/mol(1)\\
T{N_2}:\\
M + {H_2}S{O_4} \to MS{O_4} + {H_2}\\
{n_{{H_2}S{O_4}}} = 0,25 \times 0,5 = 0,125\,mol\\
\text{ Axit dư sau phản ứng } \Rightarrow {n_M} < 0,125\,mol \Rightarrow {M_M} > \dfrac{{2,4}}{{0,125}}\\
\Rightarrow {M_M} > 19,2\,g/mol(2)\\
(1),(2) \Rightarrow {M_M} = 24\,g/mol \Rightarrow {M_M} = 24\,g/mol
\end{array}\)
Vậy M là Magie
