`#Ly`

`m_(CuO` `=` `50.80%=40` $(g)$

`n_(CuO` `=` `m/M` `=` `(40)/(80)` `=` `0,5` $(mol)$

`m_(Fe_2O_3` `=` `50-40=10` $(g)$

`n_(Fe_2O_3` `=` `m/M=(10)/(160)=0,0625` $(mol)$

PTHH: 

$(1):$ $CuO$ $+$ $H_2$ $\xrightarrow[]{t^o}$ $Cu$ $+$ $H_2O$

$(2):$ $Fe_2O_3$ $+$ $3H_2$ $\xrightarrow[]{t^o}$ $2Fe$ $+$ $3H_2O$

Theo PT $(1)$, `n_(H_2` `=` `n_(CuO` `=` `0,5` $(mol)$

Theo PT $(2)$, `n_(H_2` `=` `3n_(Fe_2O_3` `=` `0,1875` $(mol)$

`=>` `∑``n_(H_2` `=` `0,5+0,1875=0,6875` $(mol)$

`V_(H_2(đktc)` `=` `n.22,4=0,6875.24,79=17,04` $(l)$

\(\begin{array}{l}
m_{CuO}=50.80\%=40(g)\\
m_{Fe_2O_3}=50-40=10(g)\\
n_{CuO}=\frac{40}{80}=0,5(mol)\\
n_{Fe_2O_3}=\frac{10}{160}=0,0625(mol)\\
CuO+H_2\xrightarrow{t^o}Cu+H_2O\\
Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O\\
Theo\,PT:\,\sum n_{H_2}=n_{CuO}+3n_{Fe_2O_3}\\
\to\sum n_{H_2}=0,5+3.0,0625=0,6875(mol)\\
\to V_{H_2(đkc)}=0,6875.24,79=17,043125(l)
\end{array}\)