Hòa tan hoàn toàn 4,05 gam oxit M2Om trong dung dịch H2SO4 10%...

$\begin{aligned}
& + \left\{ \begin{aligned}
& {{n}_{{{M}_{2}}{{O}_{m}}}}={{n}_{{{M}_{2}}{{(S{{O}_{4}})}_{m}}}}=x \\
& {{n}_{{{H}_{2}}S{{O}_{4}}}}=mx \\
\end{aligned} \right. \Rightarrow {{m}_{\text{dd} {{H}_{2}}S{{O}_{4}}}}=\dfrac{98mx}{10\%}=980mx gam\Rightarrow {{m}_{dd {{M}_{2}}{{(S{{O}_{4}})}_{m}}}}=(4,05+980mx) gam \\
& \Rightarrow \left\{ \begin{aligned}
& {{m}_{{{M}_{2}}{{O}_{m}}}}=(2M+16m)x=4,05 \\
& C{{\%}_{{{M}_{2}}{{(S{{O}_{4}})}_{m}}}}=\dfrac{(2M+96m)x}{4,05+980mx}=15,174\% \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& Mx=1,625 \\
& mx=0,05 \\
\end{aligned} \right.\Rightarrow \dfrac{M}{m}=32,5\Rightarrow \left\{ \begin{aligned}
& M=65 \\
& m=2 \\
\end{aligned} \right.\Rightarrow x=0,025 \\
& \Rightarrow {{n}_{ZnS{{O}_{4}}.n{{H}_{2}}O}}=0,05 \Rightarrow {{M}_{ZnS{{O}_{4}}.n{{H}_{2}}O}}=\dfrac{14,35}{0,05}=287\Rightarrow n=7\Rightarrow \\
\end{aligned}$