Đáp án:
\(\begin{array}{l} a,\ \%m_{Al}=57,45\%\\ \%m_{Mg}=42,55\%\\ b,\ m_{\text{dd H$_2$SO$_4$}}=63\ g.\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\ PTHH:\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑\ (1)\\ Mg+H_2SO_4\to MgSO_4+H_2↑\ (2)\\ n_{H_2}=\dfrac{1,568}{22,4}=0,07\ mol.\\ \text{Gọi $n_{Al}$ là a (mol), $n_{Mg}$ là b (mol).}\\ \text{Theo đề bài ta có hệ pt:}\ \left \{ \begin{array}{l}27a+24b=1,41\\1,5a+b=0,07\end{array} \right.\ \Rightarrow \left \{ \begin{array}{l}a=0,03\\b=0,025\end{array} \right.\\ \Rightarrow \%m_{Al}=\dfrac{0,03\times 27}{1,41}\times 100\%=57,45\%\\ \%m_{Mg}=\dfrac{0,025\times 24}{1,41}\times 100\%=42,55\%\\ b,\ Theo\ pt:\ n_{H_2SO_4}=n_{H_2}=0,07\ mol.\\ \Rightarrow V_{\text{dd H$_2$SO$_4$}}=\dfrac{0,07}{2}=0,035\ lít=35\ ml.\\ \Rightarrow m_{\text{dd H$_2$SO$_4$}}=35\times 1,8=63\ g.\end{array}\)
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