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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, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b, Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, Theo PT: \(n_{HCl}=2n_{Fe}=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(1mol\) \(2mol\) \(1mol\)
\(0,2mol\) \(0,4mol\) \(0,2mol\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(V_{H_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)
\(m_{HCl}=n.M=0,4.=14,6\left(g\right)\)
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 )
\(m_{Fe}=0,1.56=5,6g\)
\(a.Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
b. số mol của 16,8 gam Fe:
\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
Khối lượng của HCl:
\(m_{HCl}=n.M=0,6.36,5=21,9\left(g\right)\)
c.Thể tích khí Hiđro (đktc):
\(V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)
a) nFe=0,1(mol); nHCl=0,4(mol)
PTHH: Fe + 2 HCl -> FeCl2 + H2
Ta có: 0,1/1 < 0,4/2
=> Fe hết, HCl dư, tish theo nFe.
b) nH2=nFeCl2=Fe=0,1(mol)
=> V(H2,đktc)=0,1.22,4=2,24(l)
c) mFeCl2=127.0,1=12,7(g)
a) nFe=0,1(mol); nHCl=0,4(mol) PTHH: Fe + 2 HCl -> FeCl2 + H2 Ta có: 0,1/1 < 0,4/2 => Fe hết, HCl dư, tish theo nFe. b) nH2=nFeCl2=Fe=0,1(mol) => V(H2,đktc)=0,1.22,4=2,24(l) c) mFeCl2=127.0,1=12,7(g)
a,\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right);n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,15 0,3
Ta có: \(\dfrac{0,2}{1}>\dfrac{0,15}{1}\) ⇒ H2 pứ hết,Fe dư
\(V_{H_2}=3,36\left(l\right)\) (đề cho)
b, ko tính đc k/lg dd ,chỉ tính đc thể tích dd
\(V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)
a. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)
PTHH : Zn + 2HCl -> ZnCl2 + H2
0,1 0,2 0,1
b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1
\(V_{H_2}=0,1\cdot22,4=2,24l\)
\(m_{HCl}=0,2\cdot36,5=7,3g\)
Bài 1:
1) Fe + 2HCl --> FeCl2 + H2
2) \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
_____0,3--------------->0,3--->0,3
=> nH2 = 0,3.22,4 = 6,72(l)
3) mFeCl2 = 0,3.127=38,1(g)
Bài 2
1) 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
2) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
______0,2<----------------------------------0,3
=> mAl = 0,2.27 = 5,4(g)
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
=> \(\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2.36,5}{15\%}=\dfrac{146}{3}\left(g\right)\\V_{H_2}=0,1.22,4=4,48\left(l\right)\end{matrix}\right.\)