a/ \(Fe_2O_3\left(0,1\right)+3H_2\left(0,3\right)\rightarrow2Fe\left(0,2\right)+3H_2O\)
\(n_{Fe_2O_3}=\frac{16}{160}=0,1\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
\(\Rightarrow m_{Fe}=0,2.56=11,2\left(g\right)\)
b/ \(CuO\left(0,3\right)+H_2\left(0,3\right)\rightarrow Cu\left(0,3\right)+H_2O\)
\(n_{CuO}=\frac{32}{80}=0,4\left(mol\right)\)
Vì \(\frac{n_{H_2}}{1}=0,3< 0,4=\frac{n_{CuO}}{1}\) nên H2 phản ứng hết
\(\Rightarrow n_{CuO\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\)
\(\Rightarrow m_{CuO\left(dư\right)}=0,1.80=8\left(g\right)\)
\(\Rightarrow m_{Cu}=0,3.64=19,2\left(g\right)\)
\(\Rightarrow m_r=8+19,2=27,2\left(g\right)\)
