a) PTHH: Fe + 2HCl ===> FeCl2 + H2
b) Ta có: mHCl = \(200\cdot10,95\%=21,9\left(gam\right)\)
=> nHCl = \(\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
=> nH2 = \(\dfrac{1}{2}\cdot n_{HCl}=\dfrac{0,6}{2}=0,3\left(mol\right)\)
=> VH2(đktc) = \(0,3\cdot22,4=6,72\left(l\right)\)
c) Theo PT, nFe = \(\dfrac{1}{2}\cdot n_{HCl}=\dfrac{0,6}{2}=0,3\left(mol\right)\)
=> mFe = \(0,3\cdot56=16,8\left(gam\right)=a\)
d) Ta có: mdd sau pứ = 16,8 + 200 - 0,3 . 2 = 216,2 (gam)
Lại có: nFeCl2 = \(\dfrac{1}{2}\cdot n_{HCl}=0,3\left(mol\right)\)
=> mFeCl2 = 0,3 . 127 = 38,1 (gam)
=> C%(FeCl2) = \(\dfrac{38,1}{216,2}\cdot100\%=17,62\%\)
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