Gọi: nFe3O4 = x (mol) ⇒ nCuO = 2x (mol)
⇒ 232x + 80.2x = 19,6 ⇔ x = 0,05 (mol)
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe_3O_4}=0,05\left(mol\right)\\n_{CuO}=0,1\left(mol\right)\end{matrix}\right.\)
PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
____0,05____0,2___0,15 (mol)
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
_0,1___0,1__0,1 (mol)
Có: \(n_{H_2\left(banđau\right)}=0,4\left(mol\right)>0,2+0,1\)
⇒ Oxit bị khử hết, H2 dư.
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,15\left(mol\right)\\n_{Cu}=0,1\left(mol\right)\end{matrix}\right.\) và \(n_{H_2\left(dư\right)}=0,4-0,2-0,1=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,15.56}{0,15.56+0,1.64}.100\%\approx56,76\%\\\%m_{Cu}\approx43,24\%\end{matrix}\right.\)
\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
Bạn tham khảo nhé!
