$Cho_{}$ $8,4g_{}$ $Fe_{}$ $vào_{}$ $250g_{}$ $dung_{}$ $dịch_{}$ $HCl:_{}$ $a/_{}$ $C_{}$% $dung_{}$ $dịch_{}$ $HCl_{}$ $b/_{}$ $V_{H_2}$ $c/_{}$

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1. a)

   nFe = 8,4 / 56 = 0,15 (mol)

   Fe + 2HCl -> FeCl2 + H2

   0,15    0,3        0,15    0,15

   C% HCl = $\frac{0,3*36,5}{250}$ *100% = 4,38%

   b)

   VH2 = 0,15 * 22,4 = 3,36 (lít)

   C)

   mdd sau phản ứng = 8,4 + 250 – 0,15 * 2 = 258,1 (gam)

   mFeCl2 = 0,15 * 127 = 19,05 (gam)

   C%FeCl2 =  $\frac{19,05}{258,1}$ * 100% = 7,35%

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2. $n_{Fe}=8,4/56=0,15(mol)$

   $PTHH:Fe+2HCl→FeCl_2+H_2$

   $(mol)–0,15–0,3–0,15–0,15–$

   $a.C_{ddHCl}=\frac{0,3.36,5}{250}.100=4,38$(%)

   $b.V_{H_2}=0,15.22,4=3,36(l)$

   $c..C_{ddFeCl_2}=\frac{0,15.127}{250+8,4-0,15.2}.100=7,38$(%)

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