PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
a, Ta có: \(\left\{{}\begin{matrix}m_{Fe_2O_3}=20.60\%=12\left(g\right)\Rightarrow n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\\m_{CuO}=20-12=8\left(g\right)\Rightarrow n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\end{matrix}\right.\)
Theo pT: \(\left\{{}\begin{matrix}n_{Fe}=2n_{Fe_2O_3}=0,15\left(mol\right)\\n_{Cu}=n_{CuO}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
\(m_{Cu}=0,1.64=6,4\left(g\right)\)
b, Theo PT: \(n_{H_2}=3n_{Fe_2O_3}+n_{CuO}=0,325\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,325.22,4=7,28\left(l\right)\)
Bạn tham khảo nhé!
