Cho 25,2g sắt (II) tác dụng với axitclohiđric (HCl) thấy có Sắt (II)
a) viết pthh
b)hãy tính thể tích khí H\(_2\) thu đc ở Đktc
c)Hãy tính khối lượng FeCl\(_2\) tạo thành
d)Hãy tính khối lượng axitclohiđric tham gia phản ứng
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nFe = 5,6/56 = 0,1 (mol)
PTHH: Fe + 2HCl -> FeCl2 + H2
Mol: 0,1 ---> 0,2 ---> 0,1 ---> 0,1
VH2 = 0,1 . 22,4 = 2,24 (l)
mFeCl2 = 0,1 . 127 = 12,7 (g)
\(a) Fe + 2HCl \to FeCl_2 + H_2\\ b) n_{H_2} = n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ c) n_{FeCl_2} = n_{Fe} = 0,1(mol)\\ m_{FeCl_2} = 0,1.127 = 12,7(gam)\\ d) n_{HCl} = 2n_{Fe} = 0,2(mol) \Rightarrow C_{M_{HCl}} = \dfrac{0,2}{0,1} = 2M\\ e)n_{Fe_3O_4} = \dfrac{2,32}{232} = 0,01(mol)\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ 4n_{Fe_3O_4} = 0,04 < n_{H_2} = 0,1 \to H_2\ dư\\ \)
\(n_{Fe} = 3n_{Fe_3O_4} = 0,03(mol)\\ m_{Fe} = 0,03.56 = 1,68(gam)\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)
tỉ lệ: 1 : 2 : 1 : 1
n(mol) 0,2--->0,4------->0,2----->0,2
\(m_{FeCl_2}=n\cdot M=0,2\cdot\left(56+35,5\cdot2\right)=25,4\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\\ V_{H_2\left(dkc\right)}=n\cdot24,79=4,958\left(l\right)\)
a) \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
b) \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo phương trình hóa học: \(n_{H_2}=n_{Fe}=o,2\left(mol\right)\)
\(V_{H_2\left(dktc\right)}=n_{H_2}\times22,4=0,2\times22,4=4,48\left(l\right)\)
\(V_{H_2\left(dkc\right)}=n_{H_2}\times24,79=0,2\times24,79=4,96\left(l\right)\)
c) Theo phương trình hóa học: \(n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\)
\(m_{FeCl_2}=n_{FeCl_2}\times M_{FeCl_2}=0,2.127=25,4\left(g\right)\)
a: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{HCL}=\dfrac{10.95}{36.5}=0.3\left(mol\right)\)
\(n_{ZnCl_2}=n_{Zn}=\dfrac{13.6}{136}=0.1\left(mol\right)\)
Vì \(\dfrac{n_{Zn}}{1}< \dfrac{n_{HCl}}{2}\)
nên HCl dư
=>Tính theo mol của Zn
\(m_{Zn}=0.1\cdot65=6.5\left(g\right)\)
b: \(n_{H_2}=0.1\left(mol\right)\)
\(V_{H_2}=0.1\cdot22.4=2.24\left(lít\right)\)
a) \(n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
\(n_{ZnCl_2}=\dfrac{13,6}{136}=0,1\left(mol\right)\)
PTHH : Zn + 2HCl -> ZnCl2 + H2
0,1 0,1
Xét tỉ lệ \(\dfrac{0,3}{2}>\dfrac{0,1}{1}\) => HCl dư , ZnCl2 đủ
\(m_{Zn}=0,1.65=6,5\left(g\right)\)
b. \(V_{Zn}=0,1.22,4=2,24\left(l\right)\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)
ti le 1 : 2 : 1 : 1
n(mol) 0,5-->1--------->0,5------>0,5
\(m_{FeCl_2}=n\cdot M=0,5\cdot\left(56+35,5\cdot2\right)=63,5\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,5\cdot22,4=11,2\left(l\right)\)
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) \(m_{FeCl_2}=0,15.127=19,05\left(g\right)\)
\(n_{HCl}=0.2\cdot1=0.2\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(........0.2..............0.1\)
\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(n_{CuO}=\dfrac{16}{80}=0.2\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)
\(0.1.......0.1....0.1\)
\(\Rightarrow CuOdư\)
\(m_{Cu}=0.1\cdot64=6.4\left(g\right)\)
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) \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
Theo PTHH ta có: \(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=n_{MgCl_2}.M_{MgCl_2}=0,1.95=9,5\left(g\right)\)
b) Theo PTHH ta có: \(n_{H_2}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,1.22,4=2,24\left(l\right)\)
c) \(2H_2+O_2\rightarrow2H_2O\)
Theo PTHH: \(n_{H_2O}=\dfrac{0,1.2}{2}=0,1\left(mol\right)\)
\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=0,1.18=1,8\left(g\right)\)
a)mMg= 2,4/24=0,1(mol) nMgCl2=0,1.1/1=0,1 (mol) mMgCl2= 0,1.95=9,5(g) b)nH2=0,1.1/1=0,1(mol) v=n.22,4=2,24(lít
`a)PTHH:`
`Fe + 2HCl -> FeCl_2 + H_2 \uparrow`
`0,45` `0,9` `0,45` `0,45` `(mol)`
`n_[Fe]=[25,2]/56=0,45(mol)`
`b)V_[H_2]=0,45.22,4=10,08(l)`
`c)m_[FeCl_2]=0,45.127=57,15(g)`
`d)m_[HCl]=0,9.36,5=32,85(g)`