nFe= 11.2/56=0.2 mol
Fe + 2HCl --> FeCl2 + H2
0.2___0.4___________0.2
VH2= 0.2*22.4=4.48l
VddHCl= 0.4/2=0.2 l
nCuO= 48/80= 0.6 mol
CuO + H2 -to-> Cu + H2O
Bđ: 0.6____0.2
Pư: 0.2____0.2
Kt: 0.4____0
mCuO dư= 0.4*80=32g
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a) \(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)
Theo pt \(\Rightarrow n_{H_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\)
b) Theo pt
\(\Rightarrow n_{HCl}=n_{Fe}\cdot2=0,2\cdot2=0,4\left(mol\right)\)
\(\Rightarrow V_{HCl}=\frac{0,4}{2}=0,2\left(l\right)\)
c) PTHH: \(CuO+H_2\rightarrow Cu+H_2O\)
\(n_{CuO}=\frac{48}{80}=0,6\left(mol\right)\)
Lập tỉ lệ: \(\frac{0,6}{1}>\frac{0,2}{1}\)
\(\Rightarrow\) CuO dư
\(n_{CuO}\left(dư\right)=0,6-\left(\frac{0,2\cdot1}{1}\right)=0,4\left(mol\right)\)
\(\Rightarrow m_{CuO}\left(dư\right)=0,4\cdot80=32\left(g\right)\)
