Bạn tham khảo:
$1/$
$a/ C+O_2 \xrightarrow{t{o}} CO_2$(Hoá hợp)
$b/ 4Na+O_2 \xrightarrow{t{o}} 2Na_2O$ (Hoá hợp)
$c/ C_2H_4+3O_2 \xrightarrow{t{o}} 2CO_2+2H_2O$
$d/ 2KClO_3 \xrightarrow{t{o}} 2KCl+3O_2$ (Phân huỷ)
$2/$
$4Al+3O_2 \xrightarrow{t{o}} 2Al_2O_3$
$n_{O_2}=0,15(mol)$
$n_{Al}=\frac{4}{3}.0,15=0,2(mol)$
$m_{Al}=0,2.27=5,4(g)$
$3/$
$3Fe+2O_2 \xrightarrow{t{o}} Fe_3O_4$
$n_{Fe}=0,5(mol)$
$n_{O_2}=0,2(mol)$
$\frac{0,5}{3} > \frac{0,2}{2}$
$Fe >O_2$
$Fe$ dư, $O_2$ hết
$n_{Fe_3O_4}=\frac{0,2}{2}=0,1(mol)$
$m_{Fe_3O_4}=0,1.232=23,2(g)$
$n_{Fe(dư)}=0,3(mol)$
$m_{Fe(dư)}=(0,5-0,3).56=11,2(g)$