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2Al + 6HCl → 2AlCl3 + 3H2↑ (1)
Al2O3 + 6HCl → 2AlCl3 + 3H2O (2)
\(n_{H_2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
a) Theo pt1: \(n_{Al}=\frac{2}{3}n_{H_2}=\frac{2}{3}\times0,6=0,4\left(mol\right)\)
\(\Rightarrow m_{Al}=0,4\times27=10,8\left(g\right)\)
\(\Rightarrow m_{Al_2O_3}=21-10,8=10,2\left(g\right)\)
b) Theo PT1: \(n_{HCl}=3n_{Al}=3\times0,2=0,6\left(mol\right)\)
\(n_{Al_2O_3}=\frac{10,2}{102}=0,1\left(mol\right)\)
Theo Pt2: \(n_{HCl}=6n_{Al_2O_3}=6\times0,1=0,6\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,6+0,6=1,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=1,2\times36,5=43,8\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\frac{43,8}{36\%}=121,67\left(g\right)\)
\(\Rightarrow V_{ddHCl}=\frac{121,67}{1,18}=103,11\left(ml\right)\)
Bn ơi số mol của H2 là 0,6.Mà số mik của HCl là gấp 2 lần số mol của H2 là bằng 1,2 chứ.Tại sao lại bằng 0,6
Đặt nMg=a(mol); nAl=b(mol)
PTHH: Mg +2 HCl -> MgCl2 + H2
a________2a_______a_____a(mol)
2 Al + 6 HCl -> 2 AlCl3 +3 H2
b_____3b____b_____1,5b(mol)
Ta có hpt: \(\left\{{}\begin{matrix}24a+27b=7,8\\a+1,5b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
=> %mMg=[(0,1.24)/7,8].100=30,769%
=>%mAl= 69,231%
c) MgCl2 + 2 NaOH -> Mg(OH)2 + 2 NaCl
0,1_______________0,1(mol)
AlCl3 + 3 NaOH -> Al(OH)3 + 3 NaCl
0,2____________0,2(mol)
=> m=m(kết tủa)= mMg(OH)2+ mAl(OH)3= 58.0,1+ 78.0,2= 21,4(g)
\(Al+NaOH+H_2O\rightarrow NaAlO_2+\dfrac{3}{2}H_2\)
0,2 <--------------------------------------- 0,3
\(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\) (1)
0,12 <------------------------- 0,12
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (2)
0,32 --------------------------->0,48
\(n_{Al\left(2\right)}=\dfrac{16}{10}.0,2=0,32\left(mol\right)\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ \Rightarrow n_{H_2\left(1\right)}=0,6-n_{H_2\left(2\right)}=0,6-0,48=0,12\left(mol\right)\)
Trong 16 gam hỗn hợp A có:
\(m_{Al_2O_3}=16-m_{Mg}-m_{Al}=16-24.0,12-0,32.27=4,48\left(g\right)\)
\(\%_{m_{Mg}}=\dfrac{24.0,12.100\%}{16}=18\%\)
\(\%_{m_{Al}}=\dfrac{27.0,32.100\%}{16}=54\%\)
\(\%_{m_{Al_2O_3}}=\dfrac{4,48.100\%}{16}=28\%\)
Gọi \(\left\{{}\begin{matrix}n_{Mg}:a\left(mol\right)\\n_{Fe}:b\left(mol\right)\\n_{Al}:c\left(mol\right)\end{matrix}\right.\)
\(m_{O2}=51,6-32,4=19,2\left(g\right)\)
\(\Rightarrow n_{O2}=\frac{19,2}{32}=0,6\left(mol\right)\)
TN1:
\(Mg-2e\rightarrow Mg^{+2}\)
a_____2a_____
\(O_2+2e\rightarrow O^{-2}\)
0,6___1,2____
\(3Fe-1e\rightarrow Fe^{+\frac{8}{3}}\)
b____b/3_______
\(Al-3e\rightarrow Al^{+3}\)
c_____3c______
Theo BTe
\(\Rightarrow\left\{{}\begin{matrix}2a+\frac{b}{3}+3c=1,2\left(1\right)\\24a+56b+27c=32,4\left(2\right)\end{matrix}\right.\)
TN2: Gọi \(\left\{{}\begin{matrix}n_{Mg}:ka\left(mol\right)\\n_{Fe}:kb\left(mol\right)\\n_{Al}:kc\left(mol\right)\end{matrix}\right.\)
Ta có:
\(ka+kb+kc=0,9\)
\(n_{H2}=\frac{24,64}{22,4}=1,1\left(mol\right)\)
\(\Rightarrow ka+kb+1,5kc=1,1\)
\(\Rightarrow\frac{ka+kb+kc}{ka+kb+1,5kc}=\frac{0,9}{1,1}\)
\(\Rightarrow\frac{a+b+c}{a+b+1,5c}=\frac{9}{11}\)
\(\Rightarrow2a+2b-12,5c=0\left(3\right)\)
\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow\left\{{}\begin{matrix}a=0,36\\b=0,066\\c=0,117\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\frac{0,36.24.100}{32,4}=27\%\\\%m_{Fe}=\frac{0,366.56.100}{32,4}=63\%\\\%m_{Al}=100\%-27\%-63\%=10\%\end{matrix}\right.\)
TN1: Gọi \(\left(n_{Mg};n_{Al};n_{Zn}\right)=\left(a;b;c\right)\)
=> 24a + 27b + 65c = 28,6 (1)
\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: 2Mg + O2 --to--> 2MgO
a--->0,5a
4Al + 3O2 --to--> 2Al2O3
b-->0,75b
2Zn + O2 --to--> 2ZnO
c--->0,5c
=> 0,5a + 0,75b + 0,5c = 0,5 (2)
TN2: Gọi \(\left(n_{Mg};n_{Al};n_{Zn}\right)=\left(ak;bk;ck\right)\)
=> ak + bk + ck = 0,8 (3)
PTHH: Mg + 2HCl --> MgCl2 + H2
ak----------------------->ak
2Al + 6HCl -->2AlCl3 + 3H2
bk------------------------>1,5bk
Zn + 2HCl --> ZnCl2 + H2
ck---------------------->ck
=> \(ak+1,5bk+ck=\dfrac{22,4}{22,4}=1\) (4)
(1)(2)(3)(4) => \(\left\{{}\begin{matrix}a=0,2\\b=0,4\\c=0,2\\k=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2.24}{28,6}.100\%=16,783\%\\\%m_{Al}=\dfrac{0,4.27}{28,6}.100\%=37,762\%\\\%m_{Zn}=\dfrac{0,2.65}{28,6}.100\%=45,455\end{matrix}\right.\)
- Đặt \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\Rightarrow27a+24b=10,2\left(1\right)\)
Khí thu được sau p/ứ là khí H2: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
2 3 (mol)
a 3/2 a (mol)
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
1 1 (mol)
b b (mol)
Từ hai PTHH trên ta có: \(\dfrac{3}{2}a+b=0,5\left(2\right)\)
\(\left(1\right),\left(2\right)\) ta có hệ: \(\left\{{}\begin{matrix}27a+24b=10,2\\\dfrac{3}{2}a+b=0,5\end{matrix}\right.\)
Giải ra ta có \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)
a) \(\%Al=\dfrac{m_{Al}}{m_{hh}}.100\%=\dfrac{0,2.27}{10,2}.100\%\approx52,94\%\)
\(\%Mg=100\%-\%Al=100\%-52,94=47,06\%\)
b)
\(3H_2+Fe_2O_3\rightarrow^{t^0}2Fe+3H_2O\)
3 1 2 (mol)
0,5 1/6 1/3 (mol)
\(m_{Fe}=\dfrac{1}{3}.56=\dfrac{56}{3}\left(g\right)\)
\(m_{Fe_2O_3\left(pứ\right)}=\dfrac{1}{6}.160=\dfrac{80}{3}\left(g\right)\)
\(m_{Fe_2O_3\left(dư\right)}=60-m_{Fe}=60-\dfrac{56}{3}=\dfrac{124}{3}\left(g\right)\)
\(a=\dfrac{124}{3}+\dfrac{80}{3}=68\left(g\right)\)
