a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
b, \(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)
c, \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Fe}+2n_{CuO}=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\)