Đáp án:

$a,m_{NaOH}=16g.$

$b,V_{H_2}=4,48l.$

$c,m_{Fe}=6,72g.$

Giải thích các bước giải:

$a,PTPƯ:2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$

$n_{Na}=\dfrac{9,2}{23}=0,4mol.$

$Theo$ $pt:$ $n_{NaOH}=n_{Na}=0,4mol.$

$⇒m_{NaOH}=0,4.40=16g.$

$b,Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{Na}=0,2mol.$

$⇒V_{H_2}=0,2.22,4=4,48l.$

$c,PTPƯ:Fe_3O_4+4H_2\xrightarrow{t^o} 3Fe+4H_2O$

$Theo$ $pt:$ $n_{Fe}=\dfrac{3}{4}n_{H_2}=0,15mol.$

$⇒m_{Fe}=0,15.56=8,4g.$

$\text{Mà H%=80% nên:}$

$⇒m_{Fe}=8,4.80\%=6,72g.$

chúc bạn học tốt!

$n_{Na}=9,2/23=0,4mol$

$PTHH:$

$2Na+2H_2O\to 2NaOH+H_2$

a/Theo pt :

$n_{NaOH}=n_{Na}=0,4mol$

$⇒m_{NaOH}=0,4.40=16g$

b/Theo pt :

$n_{H_2}=1/2.n_{Na}=0,2mol$

$⇒V_{H_2}=0,2.22,4=4,48l$

$c/4H_2+Fe_3O_4\overset{t^o}\to 4H_2O+3Fe$

$ n_{Fe lt}=3/4.n_{H_2}=1/4.0,2=0,15mol$

$H=80\%⇒n_{Fett}=0,15.80\%=0,12mol$

$⇒m_{Fe}=0,12.56=6,72g$