Đáp án:

Giải thích các bước giải:

Đáp án:

Giải thích các bước giải:

$4/$

$a,4Al+3O_2\buildrel{{t^o}}\over\longrightarrow$ $2Al_2O_3$

$b,3Fe+2O_2\buildrel{{t^o}}\over\longrightarrow$ $Fe_3O_4$

$c,4P+5O_2\buildrel{{t^o}}\over\longrightarrow$ $2P_2O_5$

$d,CH_4+2O_2\buildrel{{t^o}}\over\longrightarrow$ $CO_2+2H_2O$

$e,2KMnO_4\buildrel{{t^o}}\over\longrightarrow$ $K_2MnO_4+MnO_2+O_2↑$

$f,2KClO_3\buildrel{{t^o}}\over\longrightarrow$ $2KCl+3O_2↑$

$g,2Al+6HCl→2AlCl_3+3H_2↑$

$5/$

$PTPƯ:3CO+Fe_2O_3\buildrel{{t^o}}\over\longrightarrow$ $2Fe+3CO_2$

$n_{Fe}=\frac{11,2}{56}=0,2mol.$

$Theo$ $pt:$ $n_{Fe_2O_3}=\frac{1}{2}n_{Fe}=0,1mol.$

$⇒m_{Fe_2O_3}=0,1.160=16g.$

$Theo$ $pt:$ $n_{CO}=\frac{3}{2}n_{Fe}=0,3mol.$

$⇒V_{CO}=0,3.22,4=6,72l.$

$6/$

$PTPƯ:3Fe+2O_2\buildrel{{t^o}}\over\longrightarrow$ $Fe_3O_4$

$n_{Fe_3O_4}=\frac{23,2}{232}=0,1mol.$

$Theo$ $pt:$ $n_{Fe}=3n_{Fe_3O_4}=0,3mol.$

$⇒m_{Fe}=0,3.56=16,8g.$

$Theo$ $pt:$ $n_{O_2}=2n_{Fe_3O_4}=0,2mol.$

$⇒V_{O_2}=0,2.22,4=4,48l.$

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