a, 

$n_{H_2}=\dfrac{2,24}{22,4}=0,1 mol$

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

$\Rightarrow n_{Na}=0,1.2=0,2 mol$

$m_{Na}=0,2.23=4,6g$

$\Rightarrow m_{Fe}=10,2-4,6=5,6g$

b,

$n_{Fe}=\dfrac{5,6}{56}=0,1 mol$

$2Na+2HCl\to 2NaCl+H_2$

$Fe+2HCl\to FeCl_2+H_2$

$\Rightarrow n_{H_2}=\dfrac{1}{2}n_{Na}+n_{Fe}=0,2 mol$

$\Rightarrow V_{H_2}=0,2.22,4=4,48l$

Đáp án:

\(a,\ m_{Na}=4,6\ g.\\ m_{Fe}=5,6\ g.\\ b,\ V_{H_2}=4,48\ lít.\)

Giải thích các bước giải:

\(a,\ PTHH:2Na+2H_2O\to 2NaOH+H_2↑\\ \text{(Fe không tan trong nước)}\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\ mol.\\ Theo\ pt:\ n_{Na}=2n_{H_2}=0,2\ mol.\\ ⇒m_{Na}=0,2\times 23=4,6\ g.\\ ⇒m_{Fe}=10,2-4,6=5,6\ g.\\ b,\ PTHH:2Na+2HCl\to 2NaCl+H_2↑\ (1)\\ Fe+2HCl\to FeCl_2+H_2↑\ (2)\\ n_{Fe}=\dfrac{5,6}{56}=0,1\ mol.\\ ⇒∑n_{H_2}=\dfrac12n_{Na}+n_{Fe}=0,1+0,1=0,2\ mol.\\ ⇒V_{H_2}=0,2\times 22,4=4,48\ lít.\)

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