Đáp án:

\(\begin{array}{l} a,\ m_{\text{dd H$_2$SO$_4$}}=196\ g.\\ b,\ m_{Fe}=5,6\ g.\\ m_{Al}=5,4\ g.\\ c,\ m_{\text{dd AgNO$_3$}}=850\ g.\\ m_{Ag}=86,4\ g.\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\ PTHH:\\ Fe+H_2SO_4\to FeSO_4+H_2↑\ (1)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑\ (2)\\ n_{H_2}=\dfrac{8,96}{22,4}=0,4\ mol.\\ Theo\ pt:\ n_{H_2SO_4}=n_{H_2}=0,4\ mol.\\ \Rightarrow m_{\text{dd H$_2$SO$_4$}}=\dfrac{0,4\times 98}{20\%}=196\ g.\\ b,\ \text{Gọi $n_{Fe}$ là a (mol), $n_{Al}$ là b (mol).}\\ \text{Theo đề bài ta có hệ pt:}\ \left \{ \begin{array}{l}56a+27b=11\\a+1,5b=0,4\end{array} \right.\ \Rightarrow \left \{ \begin{array}{l}a=0,1\\b=0,2\end{array} \right.\\ \Rightarrow m_{Fe}=0,1\times 56=5,6\ g.\\ m_{Al}=0,2\times 27=5,4\ g.\\ c,\ PTHH:\\ Al+3AgNO_3\to Al(NO_3)_3+3Ag\ (3)\\ Fe+2AgNO_3\to Fe(NO_3)_2+2Ag\ (4)\\ \Rightarrow ∑n_{AgNO_3}=3n_{Al}+2n_{Fe}=0,8\ mol.\\ \Rightarrow m_{\text{dd AgNO$_3$}}=\dfrac{0,8\times 170}{16\%}=850\ g.\\ n_{Ag}=n_{AgNO_3}=0,8\ mol.\\ \Rightarrow m_{Ag}=0,8\times 108=86,4\ g.\end{array}\)

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