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a)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,2----------->0,2----->0,3
=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b) \(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
c)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
0,3<----------------0,3
=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)
\(a,n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2--------------->0,2------->0,3
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\\ b,m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
c, PTHH:
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,2<------------------0,2
\(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)
\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,2 0,3 ( mol )
\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72l\)
\(m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,2.133,5=26,7g\)
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) PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(2mol\) \(6mol\) \(2mol\) \(3mol\)
\(0,27\) \(x\) \(y\) \(z\)
b) ta có: \(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{7,3}{27}=0,27\left(mol\right)\)
theo PT: \(n_{Al}=n_{AlCl_3}=0,27\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,27.133,5=36,045\left(g\right)\)
c) ta có: \(n_{H_2}=\dfrac{m_{H_2}}{M_{H_2}}=\) \(\dfrac{0,27.3}{2}=0,405\left(mol\right)\)
\(\Rightarrow V_{H_2\left(đktc\right)}=n_{H_2}.22,4=0,405.22,4=9,072\left(l\right)\)
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. \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b. \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,5-------1---------0,5-----0,5
Theo PTHH: \(\Rightarrow n_{H_2}=n_{Fe}=0,5\left(mol\right)\)
\(V_{H_2}=n_{H_2}.22,4=0,5.22,4=11,2\left(l\right)\)
c. \(H_2+CuO\rightarrow Cu+H_2O\)
0,5-------0,5-----0,5----0,5
\(\Rightarrow m_{Cu}=n_{Cu}.M_{Cu}=0,5.64=32\left(g\right)\)
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)
c, \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\Rightarrow m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)
d, \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Có: \(\dfrac{0,2}{1}>\dfrac{0,45}{3}\) → Fe2O3 dư.
\(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)