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$a) Fe + 2HCl \to FeCl_2 + H_2$
$b) n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)$
Theo PTHH : $n_{HCl} = 2n_{Fe} = 0,1.2 = 0,2(mol)$
$m_{HCl} = 0,2.36,5 = 7,3(gam)$
$c) n_{H_2} = n_{Fe} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(lít)$
a: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b: \(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)=n_{FeCl_2}\)
\(\Leftrightarrow n_{HCl}=2\cdot0.1=0.2\left(mol\right)\)
\(m=0.2\cdot36.5=7.3\left(g\right)\)
c: \(V_{H_2}=0.1\cdot22.4=2.24\left(lít\right)\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)
ti le 1 : 2 : 1 : 1
n(mol) 0,5-->1--------->0,5------>0,5
\(m_{FeCl_2}=n\cdot M=0,5\cdot\left(56+35,5\cdot2\right)=63,5\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,5\cdot22,4=11,2\left(l\right)\)
Bảo toàn KL: \(m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}\)
\(\Rightarrow m_{FeCl_2}=5,6+7,3-0,2=12,7(g)\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ b,n_{HCl}=2n_{H_2}=0,3(mol);n_{FeCl_2}=n_{H_2}=0,15(mol)\\ \Rightarrow m_{HCl}=0,3.36,5=10,95(g)\\ m_{FeCl_2}=0,15.127=19,05(g)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,3\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,3\cdot36,5=10,95\left(g\right)\\m_{Fe}=0,15\cdot56=8,4\left(g\right)\\m_{FeCl_2}=0,15\cdot127=19,05\left(g\right)\end{matrix}\right.\)
PTHH: Fe+2HCl→FeCl2+H2↑Fe+2HCl→FeCl2+H2↑
Ta có: nH2=3,3622,4=0,15(mol)nH2=3,3622,4=0,15(mol)
⇒{nHCl=0,3(mol)nFeCl2=nFe=0,15(mol)⇒{nHCl=0,3(mol)nFeCl2=nFe=0,15(mol) ⇒⎧⎪⎨⎪⎩mHCl=0,3⋅36,5=10,95(g)mFe=0,15⋅56=8,4(g)mFeCl2=0,15⋅127=19,05(g)⇒{mHCl=0,3⋅36,5=10,95(g)mFe=0,15⋅56=8,4(g)mFeCl2=0,15⋅127=19,05(g)
\(a)\\ Fe + 2HCl \to FeCl_2 + H_2\)
b)
\(n_{Fe} = \dfrac{22,4}{56}= 0,4(mol)\\ n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\)
Ta thấy : \(n_{Fe} > n_{H_2}\) nên Fe dư.
Theo PTHH :
\(n_{Fe\ pư} = n_{H_2} = 0,3(mol)\\ \Rightarrow m_{Fe\ pư} = 0,3.56 = 16,8(gam)\)
c)
Ta có :
\(n_{FeCl_2} = n_{H_2} = 0,3(mol)\\ \Rightarrow m_{FeCl_2} = 0,3.127 = 38,1(gam)\)
\(BTKL:n_{Fe}+n_{HCl}=n_{FeCl_2}+n_{H_2}\\ \Rightarrow n_{FeCl_2}=5,6+7,3-0,2=12,7(g)\)
a.b.c.\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 0,1 0,15 ( mol )
\(V_{H_2}=0,15.22,4=3,36l\)
\(m_{AlCl_3}=0,1.133,5=13,35g\)
d.\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,15 0,1 ( mol )
\(m_{Fe}=0,1.56=5,6g\)
a: Fe+2HCl->FeCl2+H2
0,1 0,2 0,1
b: nFe=5,6/56=0,1(mol)
=>nHCl=0,2(mol)
mHCl=0,2*36,5=7,3(g)