1/
$a,n_{NaCl}=\dfrac{2,34}{58,5}=0,04mol.$
$⇒V_{NaCl}=\dfrac{0,04}{0,1}=0,4l=400ml.$
$b,m_{dd}=400.1,12=448g.$
2/
$a,PTPƯ:Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$
$n_{Fe}=\dfrac{6,72}{56}=0,12mol.$
$n_{HCl}=\dfrac{7,3}{36,5}=0,2mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,12}{1}>\dfrac{0,2}{2}$
$⇒Fe$ $dư.$
$⇒n_{Fe}(dư)=0,12-\dfrac{0,2.1}{2}=0,02mol.$
$⇒m_{Fe}=0,02.56=1,12g.$
$b,Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1mol.$
$⇒V_{H_2}=0,1.22,4=2,24l.$
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