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\(n_{Fe}=\dfrac{28}{56}=0,5mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,5 1 0,5
\(V_{H_2}=0,5\cdot22,4=11,2l\)
\(m_{HCl}=1\cdot36,5=36,5g\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)
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) \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,5-------1---------0,5------0,5
b) \(V_{H_2}=n_{H_2}.22,4=0,5.22,4=11,2\left(l\right)\)
c) \(H_2+CuO\rightarrow Cu+H_2O\)
0,5-----0,5------0,5----0,5
Khối lượng đồng tạo thành: \(m_{Cu}=n_{Cu}.64=0,5.64=32\left(g\right)\)
a) \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
PTHH: `Fe + 2HCl -> FeCl_2 + H_2`
0,5-------------------------->0,5`
b) `V_{H_2} = 0,5.22,4 = 11,2 (l)`
c) PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
0,5---->0,5
`=> m_{Cu} = 0,5.64 = 32 (g)`
\(Fe+2HCl\underrightarrow{t^o}FeCl_2+H_2\)
\(1mol\) \(1mol\)
\(0,5mol\) \(0,5mol\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(V_{H_2}=n.22,4=0,5.22,4=11,2\left(l\right)\)
\(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)
\(1mol\) \(1mol\)
\(0,5mol\) \(0,5mol\)
\(m_{Cu}=n.M=0,5.64=32\left(g\right)\)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
______0,2_________________0,2 (mol)
b, VH2 = 0,2.22,4 = 4,48 (l)
c, Ta có: \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)
PT: \(FeO+H_2\underrightarrow{t^o}Fe+H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\), ta được H2 dư.
Theo PT: \(n_{Fe}=n_{FeO}=0,1\left(mol\right)\)
⇒ mFe = 0,1.56 = 5,6 (g)
Bạn tham khảo nhé!
a) Zn + 2HCl \(\rightarrow\)ZnCl2 + H2
b) mZn = \(\dfrac{13}{65}\)=0,2 (mol)
Zn + 2HCl \(\rightarrow\)ZnCl2 + H2
(mol) 0,2 ----------------------> 0,2
\(V_{H_2}\)= 0,2 . 22,4 = 4,48(lít)
c)\(n_{FeO}\)=\(\dfrac{7,2}{72}\)=0,1 (mol)
H2 + FeO \(\underrightarrow{t^o}\)Fe + H2O
(mol) 0,1----->0,1
mFe = 0,1 . 56 = 5,6(g)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
a, \(n_{Fe}=2n_{Fe_2O_3}=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,2.56=11,2\left(g\right)\)
b, \(n_{H_2}=3n_{Fe_2O_3}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Theo PT: \(n_{Mg}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,3.24=7,2\left(g\right)\)
a)\(n_{HCl}=\dfrac{10,65}{36,5}=0,3mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,15 0,3 0,15 0,15
b)\(V_{H_2}=0,15\cdot22,4=3,36l\)
c)\(n_{CuO}=\dfrac{16}{80}=0,2mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,2 0,15 0,15
\(m_{Cu}=0,15\cdot64=9,6g\)
Chị bấm máy tính hay vậy, em bấm: 10,65/36,5 = 213/730 :V