Đáp án:
Giải thích các bước giải:
$4/$
$a,PTPƯ:4Na+O_2\overset{t^o}\to$ $2Na_2O$
$b,n_{Na}=\dfrac{5,52}{23}=0,24mol.$
$Theo$ $pt:$ $n_{O_2}=\dfrac{1}{4}n_{Na}=0,06mol.$
$⇒V_{O_2}=0,06.22,4=1,344l.$
$c,Theo$ $pt:$ $n_{Na_2O}=\dfrac{1}{2}n_{Na}=0,12mol.$
$⇒m_{Na_2O}=0,12.62=7,44g.$
$5/$
$a,PTPƯ:3Fe+2O_2\overset{t^o}\to$ $Fe_3O_4$
$b,n_{O_2}=\dfrac{13,44}{22,4}=0,6mol.$
$Theo$ $pt:$ $n_{Fe}=\dfrac{3}{2}n_{O_2}=0,9mol.$
$⇒m_{Fe}=0,9.56=50,4g.$
$c,Theo$ $pt:$ $n_{Fe_3O_4}=\dfrac{1}{2}n_{O_2}=0,3mol.$
$⇒m_{Fe_3O_4}=0,3.232=69,6g.$
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