Đáp án:
\(\begin{array}{l} a,\ \%m_{Al}=49,1\%\\ \%m_{Fe}=50,9\%\\ b,\ V_{SO_2}=10,08\ (l).\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\\ PTHH:2Al+2NaOH+2H_2O\to 2NaAlO_2+3H_2\uparrow\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\ (mol).\\ Theo\ pt:\ n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\ (mol).\\ \Rightarrow m_{Al}=0,2\times 27=5,4\ (g).\\ \Rightarrow m_{Fe}=11-5,4=5,6\ (g).\\ \Rightarrow \%m_{Al}=\dfrac{5,4}{11}\times 100\%=49,1\%\\ \%m_{Fe}=\dfrac{5,6}{11}\times 100\%=50,9\%\\ b,\\ n_{Fe}=\dfrac{5,6}{56}=0,1\ (mol).\\ PTHH:\\ 2Al+6H_2SO_{4\ \text{(đặc, nóng)}}\xrightarrow{t^o} Al_2(SO_4)_3+3SO_2\uparrow+6H_2O\ (1)\\ 2Fe+6H_2SO_{4\ \text{(đặc, nóng)}}\xrightarrow{t^o} Fe_2(SO_4)_3+3SO_2\uparrow+6H_2O\ (2)\\ \Rightarrow \sum n_{SO_2}=\dfrac{3}{2}n_{Al}+\dfrac{3}{2}n_{Fe}=0,45\ (mol).\\ \Rightarrow V_{SO_2}=0,45\times 22,4=10,08\ (l).\end{array}\)
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