a) Gọi số mol Al, Fe là a, b (mol)=> 27a + 56b = 1,1 (1)\(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2 a----->1,5a------------------->1,5a Fe + H2SO4 --> FeSO4 + H2 b------>b---------------->b=> 1,5a + b = 0,04 (2)(1)(2) => a = 0,02 (mol); b = 0,01 (mol)=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,02.27}{1,1}.100\%=49,09\%\\\%m_{Fe}=\dfrac{0,01.56}{1,1}.100\%=50,91\%\end{matrix}\right.\)b) nH2SO4 = 1,5a + b = 0,04 (mol)=> \(C\%_{dd.H_2SO_4}=\dfrac{0,04.98}{49}.100\%=8\%\)Câu trả lời: a) Gọi số mol Al, Fe là a, b (mol)=> 27a + 56b = 1,1 (1)\(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2 a----->1,5a------------------->1,5a Fe + H2SO4 --> FeSO4 + H2 b------>b---------------->b=> 1,5a + b = 0,04 (2)(1)(2) => a = 0,02 (mol); b = 0,01 (mol)=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,02.27}{1,1}.100\%=49,09\%\\\%m_{Fe}=\dfrac{0,01.56}{1,1}.100\%=50,91\%\end{matrix}\right.\)b) nH2SO4 = 1,5a + b = 0,04 (mol)=> \(C\%_{dd.H_2SO_4}=\dfrac{0,04.98}{49}.100\%=8\%\)