a)
\(CuO+CO\underrightarrow{t^o}Cu+CO_2\)
\(FeO+CO\underrightarrow{t^o}Fe+CO_2\)
\(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)
b) A gồm CO và CO2
Ta có: \(\left\{{}\begin{matrix}n_{CO\left(A\right)}+n_{CO_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\\\overline{M}=\dfrac{28.n_{CO\left(A\right)}+44.n_{CO_2}}{n_{CO\left(A\right)}+n_{CO_2}}=\dfrac{130}{7}.2=\dfrac{260}{7}\left(g/mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{CO\left(A\right)}=0,3\left(mol\right)\\n_{CO_2}=0,4\left(mol\right)\end{matrix}\right.\) => \(\left\{{}\begin{matrix}m_{CO\left(A\right)}=0,3.28=8,4\left(g\right)\\m_{CO_2}=0,4.44=17,6\left(g\right)\end{matrix}\right.\)
c) Theo PTHH: \(n_{CO\left(pư\right)}=n_{CO_2}=0,4\left(mol\right)\)
nCO(bd) = nCO(pư) + nCO(A) = 0,7 (mol)
=> V = 0,7.22,4 = 15,68 (l)
Theo ĐLBTKL: \(m_{oxit}+m_{CO\left(pư\right)}=m_{KL}+m_{CO_2}\)
=> \(m=18,4+0,4.44-0,4.28=24,8\left(g\right)\)
