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`n_[Al]=[2,7]/27=0,1(mol)`
`2Al + 6HCl -> 2AlCl_3 + 3H_2 \uparrow`
`0,1` `0,3` `0,1` `0,15` `(mol)`
`a)V_[H_2]=0,15.22,4=3,36(l)`
`b)V_[dd HCl]=[0,3]/2=0,15(l)`
`=>C_[M_[AlCl_3]]=[0,1]/[0,15]~~0,67(M)`
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 0,3 0,1 0,15
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\\ V_{HCl}=\dfrac{0,3}{2}=0,15\left(l\right)\\ C_{M\left(AlCl_3\right)}=\dfrac{0,1}{0,15}=\dfrac{2}{3}M\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)
\(n_{HCl}=\dfrac{14,6}{36,5}=0,4mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 < 0,4 ( mol )
0,1 0,2 0,1 0,1 ( mol )
Chất dư là HCl
\(m_{HCl\left(dư\right)}=\left(0,4-0,2\right).36,5=7,3g\)
\(V_{H_2}=0,1.22,4=2,24l\)
\(m_{FeCl_2}=0,1.127=12,7g\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\
n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\\
pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
\(LTL:\dfrac{0,1}{1}< \dfrac{0,4}{1}\)
=> H2SO4 d
\(n_{H_2SO_4\left(pu\right)}=n_{Fe}=0,1\left(mol\right)\\
m_{H_2SO_4\left(d\right)}=\left(0,4-0,1\right).98=29,4g\)
\(n_{H_2}=n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\)
\(V_{H_2}=0,1.22,4=2,24l\\
m_{FeSO_4}=0,1.152=15,2g\)
a. \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(n_{Fe}=\frac{11,2}{56}=0,2mol\)
b. Theo phương trình \(n_{HCl}=n_{Fe}.2=0,2.2=0,4mol\)
\(\rightarrow V_{ddHCl}=\frac{0,4}{2}=0,2l=200ml\)
c. Theo phương trình \(n_{FeCl_2}=n_{Fe}=0,2mol\)
\(\rightarrow C_{M_{ddFeCl_2}}=\frac{0,2}{0,2}=1M\)
Sửa đề: 8,4 gam Fe
\(a,n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ n_{HCl}=\dfrac{14,6.175}{36,5.100}=0,7\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
ban đầu 0,15 0,7
phản ứng 0,15 0,3
sau pư 0 0,4 0,15 0,15
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
\(b,m_{dd}=8,4+175-0,15.2=183,1\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,15.127}{183,1}.100\%=10,4\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,4.36,5}{183,1}.100\%=7,97\%\end{matrix}\right.\)
a)
\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
\(m_{HCl}=\dfrac{175.14,6}{100}=25,55\left(g\right)\\ \rightarrow n_{HCl}=\dfrac{25,55}{35,5}=0,7\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
bđ 0,3 0,7
pư 0,3 0,6
spư 0 0,1 0,3 0,3
=> VH2 = 0,3.22,4 = 6,72 (l)
b)
mdd = 16,8 + 175 - 0,3.2 = 191,2 (g)
=> \(\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,3.127}{191,2}.100\%=19,93\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{191,2}.100\%=1,91\%\end{matrix}\right.\)
\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ n_{HCl}=\dfrac{\dfrac{175.14,6}{100}}{36,5}=0,7\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(LTL:\dfrac{0,3}{1}< \dfrac{0,7}{2}\)
\(n_{H_2}=n_{Fe}=0,3\left(mol\right)\\
V_{H_2}=0,3.22,4=6,72\left(l\right)\\
m_{\text{dd}}=16,8+175-\left(0,3.2\right)=191,2\left(g\right)\\
n_{FeCl_2}=n_{Fe}=0,3\left(mol\right)\\
C\%_{FeCl_2}=\dfrac{0,3.127}{191,2}.100\%=19,92\%\)
=> HCl dư
Fe + 2HCl $\to$ FeCl2 + H2
n H2 = n Fe = 2,8/56 = 0,05(mol)
=> m tăng = m Fe - m H2 = 2,8 - 0,05.2 = 2,7 gam
Vậy sau phản ứng, khối lượng dung dịch HCl tăng 2,7 gam
\(n_{Fe}=\dfrac{2.8}{56}=0.05\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0.05 0.1 (mol)
\(m_{HCl}=0.1\cdot36.5=3.65\left(g\right)\)
\(n_{Al}=\dfrac{4.5}{27}=\dfrac{1}{6}\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(\dfrac{1}{6}.....0.5.......\dfrac{1}{6}.......0.25\)
\(m_{HCl}=0.5\cdot36.5=18.25\left(g\right)\)
\(m_{AlCl_3}=\dfrac{1}{6}\cdot133.5=22.25\left(g\right)\)
\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)
n Fe = 5,6/56 = 0,1 mol
Fe + 2HCl → FeCl2 + H2
Theo PTHH :
n HCl = 2n Fe = 0,1.2 = 0,2(mol)
=> V dd = V dd HCl = 0,2/2 = 0,1(lít)
n FeCl2 = n Fe = 0,1(mol)
=> CM FeCl2 = 0,1/0,1 = 1M