a)
\(m_{CuO}=\dfrac{12.33,\left(3\right)}{100}=4\left(g\right)\) => \(m_{Fe_2O_3}=12-4=8\left(g\right)\)
X gồm \(\left\{{}\begin{matrix}CuO:4\left(g\right)\\Fe_2O_3:8\left(g\right)\end{matrix}\right.\)
\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\); \(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
0,05-->0,05------>0,05
Fe2O3 + 3H2 --to--> 2Fe + 3H2O
0,05--->0,15------>0,1
=> \(Y\left\{{}\begin{matrix}m_{Cu}=0,05.64=3,2\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)
b) \(n_{H_2}=0,05+0,15=0,2\left(mol\right)\)
=> \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)