\(\ n_{CO_2}=\dfrac{3,36}{22,4}=0,15(mol) \\n_{O_2}=\dfrac{8}{32}=0,25(mol) \\C+O_2\xrightarrow{t^o}CO_2 \\\text{Theo phương trình} \\n_{C}=n_{O_2\hspace{0,1cm}(pứ)}=n_{CO_2}=0,15(mol) \\a, \\m_{C}=0,15.12=1,8(g) \\b, \\n_{O_2\hspace{0,1cm}(dư)}=0,25-0,15=0,1(mol) \\m_{O_2\hspace{0,1cm}(dư)}=0,1.32=3,2(g) \\c, \\O_2+2CO\xrightarrow{t^o}2CO_2 \\\text{Theo phương trình} \\n_{CO}=2n_{O_2}=0,2(mol) \\\Rightarrow m_{CO}=0,2.28=5,6(g)\)