\(a.Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ b.Đặt:\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\ Tacó:\left\{{}\begin{matrix}24x+27y=10,2\\96x+342.\dfrac{y}{2}=46,2\end{matrix}\right.\\ \Rightarrow x=0,575,y=-\dfrac{2}{15}\)
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