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\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)
ti le 1 : 2 : 1 : 1
n(mol) 0,5-->1--------->0,5------>0,5
\(m_{FeCl_2}=n\cdot M=0,5\cdot\left(56+35,5\cdot2\right)=63,5\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,5\cdot22,4=11,2\left(l\right)\)
a,\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,2 0,4 0,2
b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.2.......0.4....................0.2\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)
PTHH : \(Fe+2HCl-->FeCl_2+H_2\uparrow\) (1)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
Từ (1) => \(n_{Fe}=n_{H_2}=0.2\left(mol\right)\)
=> \(V_{H2\left(đktc\right)}=n.22,4=0,2.22,4=4,48\left(l\right)\)
Từ (1) => \(2n_{Fe}=n_{HCl}=0.4\left(mol\right)\)
=> \(m_{HCl}=n.M=0,4.\left(1+35.5\right)=14.6\left(g\right)\)
a) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
_____0,1--->0,2------->0,1---->0,1
=> mHCl = 0,2.36,5 = 7,3(g)
b) mMgCl2 = 0,1.95 = 9,5 (g)
c) VH2 = 0,1.22,4 = 2,24(l)
a) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
_____0,1--->0,2------->0,1---->0,1
=> mHCl = 0,2.36,5 = 7,3(g)
b) mMgCl2 = 0,1.95 = 9,5 (g)
c) VH2 = 0,1.22,4 = 2,24(l)
a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,15<--0,3<-----0,15<--0,15
=> \(m_{Fe}=0,15.56=8,4\left(g\right)\)
=> \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
b) \(m_{FeCl_2}=0,15.127=19,05\left(g\right)\)
Fe + 2HCl -> FeCl2 + H2
nFe = 5,6/56 = 0,1 mol
=>nH2 = 0,1 mol
=> VH2= 0,1*22,4= 2,24 lít
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
0,1-->0,2------------------>0,1
=> \(\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2.36,5}{15\%}=\dfrac{146}{3}\left(g\right)\\V_{H_2}=0,1.22,4=4,48\left(l\right)\end{matrix}\right.\)
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)
\(a,n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl}=2.n_{Fe}=2.0,1=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ b,n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)