\(26,n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2SO_4}=n_{H_2}=0,2\left(mol\right)\\ \Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,2}{0,2}=1M\left(A\right)\)
\(27,PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
Vậy chọn B