\(a,n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Tỉ.lệ.p.ứ.hh=1:2:1:1\\ b,n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c,n_{HCl}=2.0,1=0,2\left(mol\right)\)
a. \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH : Fe + 2HCl -> FeCl2 + H2
Tỉ lệ : 1 : 2 : 1 :1
Số mol : 0,1 0,2 0,1 0,1
b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
n Fe=\(\dfrac{5,6}{56}=0,1mol\)
Fe+2HCl->FeCl2+H2
0,1-0,2----------------0,1 mol
tỉ lệ 1:2:1:1
=>VH2=0,1.22,4=2,24l
=>m HCl=0,2 mol
\(Fe+2HCl \rightarrow FeCl_2+H_2 \uparrow\\n_{Fe}=\dfrac{m}M=\dfrac{5,6}{56}=0,1(mol)\\\Rightarrow n_{H_2}=n_{Fe}=0,1(mol)\\\Rightarrow V_{H_2}(đktc)=0,1.22,4=2,24(lít)\)
\(a)\)Tỉ lệ phản ứng : \(1-2-1-1\)
\(b)V_{H_2}=2,24(lít)\\c)n_{HCl}=2n_{Fe}=0,1.2=0,2(mol)\)