a. Theo đề, ta có: \(n_{SO_2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
Gọi \(a,b\) lần lượt là số mol của \(Fe,Al\) .
\(\Rightarrow m_{hh}=56a+27b=15,15\left(g\right)\left(1\right)\)
PTHH:
\(2Fe+6H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
\(a------------->\frac{3}{2}a\)
\(2Al+6H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)
\(b------------->\frac{3}{2}b\)
Theo phương trình: \(n_{SO_2}=\frac{3}{2}a+\frac{3}{2}b=0,6\left(mol\right)\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\) ta có hệ: \(\left\{{}\begin{matrix}56a+27b=15,15\\\frac{3}{2}a+\frac{3}{2}b=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,15\left(mol\right)\\b=0,25\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\Rightarrow\%m_{Fe}=\frac{8,4}{15,15}.100\%=55,45\%\)
\(\Rightarrow\%m_{Al}=100\%-55,45\%=44,55\%\)
b. Cũng theo phương trình:\(n_{H_2SO_4}=3a+3b=3.0,15+3.0,25=1,2\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=1,2.98=117,6\left(g\right)\)
\(\Rightarrow C\%_{ddH_2SO_4}=\frac{117,6}{150}.100\%=78,4\%\)
