a) \(n_{Fe}=\frac{m}{M}=\frac{22,4}{56}=0,4\left(mol\right)\)
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow_{\left(1\right)}\)
Theo pt (1) ta có \(n_{FeCl_2}=n_{Fe}=0,4\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=n.M=0,4.127=50,8\left(g\right).\)
b) Theo pt (1) ta có \(n_{Fe}=n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow V_{H_2}=n.22,4=0,4.22,4=8,96\left(l\right).\)
c) \(n_{CuO}=\frac{m}{M}=\frac{8}{80}=0,1\left(mol\right).\)
\(PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O.\)
Ta có tỉ lệ: \(\frac{0,4}{1}>\frac{0,1}{1}\Rightarrow\) Cu hết, H2 dư.
\(\Rightarrow\) \(n_{Cu}=n_{CuO}\)(hết) = 0,1 (mol).
\(\Rightarrow m_{Cu}=n.M=0,1.64=6,4\left(mol\right).\)