$a,PTPƯ:Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$
$n_{Fe}=\dfrac{2,8}{56}=0,05mol.$
$n_{HCl}=\dfrac{14,6.10\%}{36,5}=0,04mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,05}{1}>\dfrac{0,04}{2}$
$⇒Fe$ $dư.$
$⇒n_{Fe}(dư)=0,05-\dfrac{0,04.1}{2}=0,03mol.$
$⇒m_{Fe}(dư)=0,03.56=1,68g.$
$b,Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{HCl}=0,02mol.$
$⇒V_{H_2}=0,02.22,4=0,448l.$
$c,n_{HCl}(thêm)=0,05.2-0,04=0,06mol.$
$⇒m_{HCl}(thêm)=0,06.36,5=2,19g.$
chúc bạn học tốt!