a) PTHH: Fe + 2HCl -> FeCl2 + H2
Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PTHH và đb, ta có:
\(n_{HCl}=2.0,2=0,4\left(mol\right)\)
=> \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)
b) PTHH: H2 + CuO -to-> Cu + H2O
Ta có: \(n_{H_2}=0,2\left(mol\right)\)
Theo PTHH va đb, ta có:
\(n_{Cu}=n_{H_2}=0,2\left(mol\right)\\ =>m_{Cu}=0,2.64=12,8\left(g\right)\)