\(n_{CO2}=n_{CaCO3}=\frac{5,5}{100}=0,055\left(mol\right)\)
\(2CuO+C\underrightarrow{^{to}}2Cu+CO_2\)
\(2PbO+C\underrightarrow{^{to}}2Pb+CO_2\)
Đặt a là mol CuO, b là mol PbO
Ta có hệ:
\(80a+223b=10,23;0,5a+0,5b=0,055\)
\(\rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,01\end{matrix}\right.\)
\(\%_{CuO}=\frac{80.0,1.100}{10,23}=78,2\%\)
\(\%PbO=100\%-78,2\%=21,8\%\)