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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
=> \(\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2.36,5}{15\%}=\dfrac{146}{3}\left(g\right)\\V_{H_2}=0,1.22,4=4,48\left(l\right)\end{matrix}\right.\)
\(n_{Fe}=\dfrac{16.8}{56}=0.3\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.3....................0.3.........0.3\)
\(m_{FeCl_2}=0.3\cdot127=38.1\left(g\right)\)
\(V_{H_2}=0.6\cdot22.4=6.72\left(l\right)\)
\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)
\(.......0.3...0.3\)
\(m_{Cu}=0.3\cdot64=19.2\left(g\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
a, Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
Theo PT: \(n_{FeCl_2}=n_{Fe}=0,3\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=0,3.127=38,1\left(g\right)\)
b, Theo PT: \(n_{H_2}=n_{Fe}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PT: \(n_{Cu}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,3.64=19,2\left(g\right)\)
Bạn tham khảo nhé!
a.b.
\(n_{Fe}=\dfrac{2,8}{56}=0,05mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,05 0,1 0,05 ( mol )
\(V_{H_2}=0,05.22,4=1,12l\)
\(m_{HCl}=0,1.36,5=3,65g\)
c.
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,05 0,05 ( mol )
\(m_{Cu}=0,05.64=3,2g\)
\(n_{Fe}=\dfrac{28}{56}=0,5mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,5 1 0,5
\(V_{H_2}=0,5\cdot22,4=11,2l\)
\(m_{HCl}=1\cdot36,5=36,5g\)
\(a)\\ Fe + 2HCl \to FeCl_2 + H_2\)
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 :
\(n_{FeCl_2} = n_{H_2} = 0,3(mol)\\ \Rightarrow m_{FeCl_2} = 0,3.127 = 38,1(gam)\)
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.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 )
\(m_{Fe}=0,1.56=5,6g\)
1) PT: Fe + 2HCl →FeCl2 +H2↑
nHCl=\(\dfrac{3,65}{36,5}\)=0,1(mol)
Theo PT ta có:n\(_{H_2}\)=\(\dfrac{1}{2}\)nHCl=\(\dfrac{1}{2}\).0,1=0,05(mol)⇒V\(_{H_2}\)=0,05.22,4=1,12(l)
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