a/ \(Fe\left(0,5\right)+2HCl\left(1\right)\rightarrow FeCl_2+H_2\left(0,5\right)\)
\(CuO\left(0,25\right)+2HCl\left(0,5\right)\rightarrow CuCl_2+H_2O\)
\(n_{Fe}=\frac{28}{56}=0,5\left(mol\right)\)
\(n_{CuO}=\frac{20}{80}=0,25\left(mol\right)\)
\(\Rightarrow n_{HCl\left(pứ\right)}=1+0,5=1,5\left(mol\right)\)
\(n_{HCl}=2.2=4\left(mol\right)\)
Vì \(n_{HCl\left(pứ\right)}=1,5< 4=n_{HCl}\) nên A tan hết.
b/ \(H_2\left(0,5\right)+FeO\left(0,5\right)\rightarrow Fe\left(0,5\right)+H_2O\left(0,5\right)\)
\(n_{FeO}=\frac{72}{72}=1\left(mol\right)\)
Vì \(n_{H_2}=0,5< 1=n_{FeO}\) nên FeO dư
\(m_{FeO\left(pứ\right)}=0,5.72=36\left(g\right)\)
\(\Rightarrow m_{FeO\left(dư\right)}=72-36=36\left(g\right)\)
Nên FeO đã giảm 36 g
\(\Rightarrow m_{Fe}=0,5.56=28\left(g\right)\)
\(m_{H_2O}=0,5.18=9\left(g\right)\)
\(\Rightarrow\%FeO=\frac{36}{36+28+9}=49,32\%\)
\(\%Fe=\frac{28}{36+28+9}.100\%=38,36\%\)
\(\Rightarrow\%H_2O=100\%-49,32\%-38,36\%=12,32\%\)
