\(n_{CO_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(C+O_2\underrightarrow{^{^{t^0}}}CO_2\)
\(0.4...........0.4\)
\(n_{C\left(bđ\right)}=\dfrac{0.4}{60\%}=\dfrac{2}{3}\left(mol\right)\)
\(m_C=\dfrac{2}{3}\cdot12=8\left(g\right)\)
\(m_{than}=\dfrac{8}{80\%}=10\left(g\right)\)