\(a,n_{O\left(bđ\right)}=n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)
BTNT: \(m_{O\left(pư\right)}=32-27,2=4,8\left(g\right)\)
\(\rightarrow n_{O\left(pư\right)}=\dfrac{4,8}{16}=0,3\left(mol\right)\)
So sánh: 0,3 < 0,4 \(\rightarrow\) trong X gồm Cu và CuO
PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
Theo PTHH: \(n_{CuO\left(pư\right)}=n_{Cu}=n_{O\left(pư\right)}=n_{H_2}=0,3\left(mol\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,3.80}{27,2}.100\%=88,24\%\\\%m_{Cu}=100\%-88,24\%=11,76\%\end{matrix}\right.\)
\(b,V_{H_2}=0,3.22,4=6,72\left(l\right)\)
\(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\\
pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
0,4 0,4 0,4
\(m_{Cu}=0,4.64=25,6\left(g\right)\\
\%m_{Cu}=\dfrac{25,6}{27,2}.100\%=94,117\%\\
\%m_{CuO\left(d\right)}=100\%-94,117\%5,883\%\)
\(V_{H_2}=0,4.22,4=8,96\left(l\right)\)
