a) Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\\n_{HCl}=0,125\cdot2=0,25\left(mol\right)\end{matrix}\right.\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,075___0,15_________0,075 (mol)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,05____0,1 (mol)
Từ PTHH: \(x=m_{Fe}+m_{CuO}=0,075\cdot56+0,05\cdot80=8,2\left(g\right)\)
b) PTHH: \(2Fe+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Fe_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
Theo PTHH: \(\left\{{}\begin{matrix}n_{H_2SO_4}=3n_{Fe}+n_{CuO}=0,275\left(mol\right)\\n_{SO_2}=0,1125\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=0,0375\left(mol\right)\\n_{CuSO_4}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,275\cdot98}{55\%}=49\left(g\right)\\m_{SO_2}=0,1125\cdot64=7,2\left(g\right)\\m_{Fe_2\left(SO_4\right)_3}=0,0375\cdot400=15\left(g\right)\\m_{CuSO_4}=0,05\cdot160=8\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{hh}+m_{ddH_2SO_4}-m_{SO_2}=50\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{15}{50}\cdot100\%=30\%\\C\%_{CuSO_4}=\dfrac{8}{50}\cdot100\%=16\%\end{matrix}\right.\)
