a, \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: Cu + H2SO4 l → ko phản ứng
PTHH: Fe + H2SO4 → FeSO4 + H2
Mol: 0,05 0,05 0,05
\(m_{Fe}=0,05.56=2,8\left(g\right)\Rightarrow m_{Cu}=3,44-2,8=0,64\left(g\right)\)
b, \(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\)