\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl}=0,05.2=0,1\left(mol\right)\\ a.V_{ddHCl}=\dfrac{0,1}{2}=0,05\left(l\right)\\ b.n_{H_2}=n_{FeCl_2}=n_{Fe}=0,05\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ c.C_{MddFeCl_2}=\dfrac{0,05}{0,05}=1\left(M\right)\)
Bài 2 :
\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,05 0,1 0,05 0,05
a) \(n_{HCl}=\dfrac{0,05.2}{1}=0,1\left(mol\right)\)
\(V_{ddHCl}=\dfrac{0,1}{2}=0,05\left(l\right)\)
b) \(n_{H2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)
c) \(n_{FeCl2}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
\(C_{M_{FeCl2}}=\dfrac{0,05}{0,05}=1\left(M\right)\)
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