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\(n_{H_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
_________\(0,2\)____\(0,6\)_____\(0,2\)______\(0,3\left(mol\right)\)
a) \(m_{Al}=0,2.27=5,4\left(g\right)\)
b) \(m_{HCl}=0,6.36,5=21,9\left(g\right)\)
\(C\%_{HCl}=\frac{21,9}{200}.100\%=10,95\%\)
c) \(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
\(C\%_{AlCl_3}=\frac{26,7}{200}.100\%=13,35\%\)
d) \(V_{AlCl_3}=\frac{26,7}{1,1}=\frac{267}{11}\left(l\right)\)
\(C_{M_{AlCl_3}}=\frac{0,2}{\frac{267}{11}}=0,007\left(M\right)\)
nCaO=0,4 mol
mH2O=1g=>nH2O=1/18mol
PTHH: CaO+H2O=> Ca(OH)2
0,4:1/18 => nCaO dư theo nH2O
Cm=1/18:1=1/18M
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Mg}=0,1\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,1.36,5}{10\%}=36,5\left(g\right)\)
c, \(n_{H_2}=n_{MgCl_2}=n_{Mg}=0,05\left(mol\right)\)
Ta có: m dd sau pư = 1,2 + 36,5 - 0,05.2 = 37,6 (g)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,05.95}{37,6}.100\%\approx12,63\%\)
Để xem rõ hơn bạn ấn chuột phải , lưu hình ảnh thành rồi coi nhá
2Al + 6HCl \(\rightarrow\)2AlCl3 + 3H2 (1)
ZnO + 2HCl \(\rightarrow\)ZnCl2 + H2O (2)
nH2=\(\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
Theo PTHH 1 ta có:
\(\dfrac{2}{3}\)nH2=nAl=0,4(mol)
mAl=0,4.27=10,8(g)
%mAl=\(\dfrac{10,8}{27}.100\%=40\%\)
%mZnO=100-40=60%
b;mZnO=27-10,8=16,2(g)
nZnO=\(\dfrac{16,2}{81}=0,2\left(mol\right)\)
Theo PTHH 1 và 2 ta có:
3nAl=nHCl(1)=1,2(mol)
2nZnO=nHCl(2)=0,4(mol)
mHCl=36,5.(0,4+1,2)=58,4(g)
mdd HCl=\(58,4:\dfrac{29,2}{100}=200\left(g\right)\)
c;
Theo PTHH 1 và 2 ta có:
nAl=nAlCl3=0,4(mol)
mAlCl3=0,4.133,5=53,4(g)
nZn=nZnCl2=0,2(mol)
mZnCl2=0,2.136=27,2(g)
C% dd AlCl3=\(\dfrac{53,4}{27+200-0,6.2}.100\%=23,65\%\)
C% dd ZnCl2=\(\dfrac{27,2}{27+200-1,2}.100\%=12,04\%\)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
de: 0,4 \(\leftarrow\) 1,2 \(\leftarrow\) 0,4 \(\leftarrow\) 0,6
\(m_{Al}=0,4.27=10,8g\)
\(m_{ZnO}=27-10,8=16,2g\)
\(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)
a, \(\%m_{Al}=\dfrac{10,8}{27}.100\%=40\%\)
\(\%m_{ZnO}=100-40=60\%\)
ZnO + 2HCl \(\rightarrow\) ZnCl2 + H2O
de: 0,2 \(\rightarrow\) 0,4 \(\rightarrow\) 0,2
b, \(m_{HCl}=36,5.\left(1,2+0,2\right)=51,1g\)
\(m_{ddHCl}=\dfrac{51,1}{29,2}.100=175g\)
c, \(m_{ZnCl_2}=0,2.136=27,2g\)
\(m_{AlCl_3}=0,4.133,5=53,4g\)
\(m_{dd}=175+27-0,6.2=200,8g\)
\(C\%_{ZnCl_2}=\dfrac{27,2}{200,8}.100\%\approx13,55\%\)
\(C\%_{AlCl_3}=\dfrac{53,4}{200,8}.100\%\approx26,59\%\)
Bài 3:
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Gọi x, y lần lượt là số mol của Mg, Al
PTHH:
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
...x.............................x.............x
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
..y.............................y...........1,5y
Ta có hệ PT: \(\left\{{}\begin{matrix}24x+27y=7,8\\x+1,5y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
=> \(m_{Mg}=0,1.24=2,4\left(g\right)\)
a. => \(\%Mg=\dfrac{2,4}{7,8}.100\%=30,77\%\)
=> \(\%Al=100\%-30,76\%=69,23\%\)
b. \(m_{MgCl_2}=0,1.95=9,5\left(g\right)\)
\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
\(\Rightarrow m_{muoi-khan}=9,5+26,7=36,2\left(g\right)\)
nH2 = 0,75 mol
2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
0,5<---1,5<----------0,5<--0,75
\(\Rightarrow\) mAl = 0,5.27 = 13,5 (g)
\(\Rightarrow\) VHCl = \(\dfrac{1,5}{3}\) = 0,5 (l)
\(\Rightarrow\) mAlCl3 = 0,5.133,5 = 66,75 (g)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\\ a,m_{AlCl_3}=133,5.0,1=13,35\left(g\right)\\ n_{H_2}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ c,n_{HCl}=\dfrac{6}{2}.0,1=0,3\left(mol\right)\\ c,C_{MddHCl}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
\(a.n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 0,3 0,1 0,15
\(m_{Al}=0,1.27=2,7g\\ b.C_{M_{HCl}}=\dfrac{0,3}{0,3}=1M\\ c.C_{\%HCl}=\dfrac{0,3.36,5}{300.1,2}\cdot100=3,04\%\\ d)m_{dd}=2,7+300.1,2-0,15.2=362,4g\\ C_{\%AlCl_3}=\dfrac{0,1.133,5}{362,4}\cdot100=3,68\%\)