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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
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\\ V_{HCl}=\dfrac{0,3}{2}=0,15\left(l\right)\\ C_{M\left(AlCl_3\right)}=\dfrac{0,1}{0,15}=\dfrac{2}{3}M\)
\(a)n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ 2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\\ n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\cdot0,4=0,6\left(mol\right)\\ V_{H_2}=0,6.22,4=13,44\left(l\right)\\ b)n_{HCl}=3n_{Al}=3.0,4=1,2\left(mol\right)\\ m_{HCl}=1,2.36,5=43,8\left(g\right)\\ m_{dd_{HCl}}=\dfrac{43,8}{10,95\%}\cdot100\%=400\left(g\right)\\ c)n_{AlCl_3}=n_{Al}=0,4mol\\ m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\\ m_{H_2}=0,6.2=1,2\left(g\right)\\ m_{dd_{AlCl_3}}=10,8+400-1,2=409,6\left(g\right)\\ C_{\%AlCl_3}=\dfrac{53,4}{409,6}\cdot100\%\approx13\%\)
\(n_{Al}=\frac{10,8}{27}=0,4\left(mol\right)\)
\(2Al+6HCl->2AlCl_3+3H_2\) (1)
theo (1) \(n_{H_2}=\frac{3}{2}n_{Al}=0,6\left(mol\right)\)
=> \(V_{H_2}=0,6.22,4=13,44\left(l\right)\)
Cho 10,08 g nhom tac dung vua du voi dung dich axit HCl.2M
a) viet phuong trinh phan ung va tinh the tich H2(dktc)
b) tinh the tich dung dich axit HCl.2M da dung
nAl = 5.4 / 27 = 0.2 (mol)
2Al + 6HCl => 2AlCl3 + 3H2
0.2......0.6............0.2.......0.3
a) VH2 = 0.3 * 22.4 = 6.72 (l)
b) mAlCl3 = 0.2 * 133.5 = 26.7 (g)
c) VddHCl = 0.6 / 1.5 = 0.4 (l)
d) CMAlCl3 = 0.2 / 0.4 = 0.5 (M)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{HCl}=\dfrac{0,6}{1,5}=0,4\left(l\right)=400\left(ml\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\end{matrix}\right.\)
\(n_{HCl}=0,3.2=0,6\left(mol\right)\\a, 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b,n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,n_{Al}=n_{AlCl_3}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ m_{Al}=0,2.27=5,4\left(g\right)\\ d,V_{ddAlCl_3}=V_{ddHCl}=0,3\left(l\right)\\ C_{MddHCl}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)
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
nMg = 6,72 : 22,4 = 0,3 mol
Mg + 2HCl -> MgCl2 + H2
0,3 0,6 0,3
=> mMg = 0,3 . 24 = 7,2 g
CM HCl = 0,6 : 0,5 = 4M
\(n_{Al}=\dfrac{4.5}{27}=\dfrac{1}{6}\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(\dfrac{1}{6}.....0.5.......\dfrac{1}{6}.......0.25\)
\(m_{HCl}=0.5\cdot36.5=18.25\left(g\right)\)
\(m_{AlCl_3}=\dfrac{1}{6}\cdot133.5=22.25\left(g\right)\)
\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)
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