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a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Al}:x\left(mol\right)\\n_{Fe}:y\left(mol\right)\end{matrix}\right.\)
Ta có : \(\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\)
=> x=0,2 ; y=0,1
\(\%m_{Al}=\dfrac{0,2.27}{11}.100==49,09\%\)
\(\%m_{Fe}=50,91\%\)
b) \(\Sigma n_{HCl}=3x+2y=0,8\left(mol\right)\)
=> \(V_{HCl}=\dfrac{0,8}{2}=0,4\left(lít\right)\)
c) \(CM_{AlCl_3}=\dfrac{0,2}{0,4}=0,5M\)
\(CM_{FeCl_2}=\dfrac{0,1}{0,4}=0,25M\)
\(a)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\
Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2 0,1 0,1
\(n_{Fe_2O_3}=\dfrac{21,6-56.0,1}{160}=0,1mol\\
Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2\)
0,1 0,6 0,2 0,3
\(V_{ddHCl}=\dfrac{0,2+0,6}{1}=0,8l\\
b.C_{M_{FeCl_2}}=\dfrac{0,1}{0,8}=0,125M\\
C_{M_{FeCl_3}}=\dfrac{0,2}{0,8}=0,25M\)
a)
Gọi $n_{Fe} = a(mol) ; n_{Al} =b (mol) \Rightarrow 56a + 27b = 11(1)$
$Fe + 2HCl \to FeCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : $n_{H_2} = a + 1,5b = \dfrac{8,96}{22,4} = 0,4(2)$
Từ (1)(2) suy ra : a = 0,1 ; b = 0,2
$\%m_{Fe} = \dfrac{0,1.56}{11}.100\% = 50,9\%$
$\%m_{Al} = 100\% - 50,9\% = 49,1\%$
b) $n_{HCl} = 2n_{H_2} = 0,8(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,8}{0,4} = 2M$
c)
$C_{M_{FeCl_2}} = \dfrac{0,1}{0,4} = 0,25M$
$C_{M_{AlCl_3}} =\dfrac{0,2}{0,4} = 0,5M$
\(n_{H_2}= \dfrac{8,96}{22,4} = 0,4(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ m_{tăng} = m_{kim\ loại} - m_{H_2} = 11 - 0,4.2 = 10,2(gam)\)
PTHH: 2Al+6HCl→2AlCl3+3H2↑2Al+6HCl→2AlCl3+3H2↑
Fe+2HCl→FeCl2+H2↑Fe+2HCl→FeCl2+H2↑
Ta có: mH2=8,9622,4⋅2=0,8(g)<mhh=11(g)mH2=8,9622,4⋅2=0,8(g)<mhh=11(g)
⇒⇒ Sau p/ứ dd tăng 11−0,8=10,2(g)
\(a)2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b)n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\\ n_{Al}=a;n_{Fe}=b\\ \left\{{}\begin{matrix}3a+b=0,25\\27a+56b=8,3\end{matrix}\right.\\ a=\dfrac{19}{470};b=\dfrac{121}{940}\\ \%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{8,3}\cdot100=13,15\%\\ \%m_{Fe}=100-13,15=86,85\%\\ c)n_{HCl}=3\cdot\dfrac{19}{470}+2\cdot\dfrac{121}{940}=\dfrac{89}{235}mol\\ m_{ddHCl=}=\dfrac{\dfrac{89}{235}\cdot36,5}{7,3}\cdot100=189g\\ d)n_{AlCl_3}=n_{Al}=\dfrac{19}{470}mol\\ n_{Fe}=n_{FeCl_2}=\dfrac{121}{940}mol\)
\(m_{dd}=8,3+189-0,25.2=196,8g\\ C_{\%AlCl_3}=\dfrac{\dfrac{19}{470}\cdot133,8}{196,8}\cdot100=2,8\%\\ C_{\%FeCl_2}=\dfrac{\dfrac{121}{940}127}{196,8}\cdot100=8,3\%\)
\(\text{Đặt }n_{Al}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 27x+56y=13,9(1)\\ n_{H_2}=\dfrac{7,84}{22,4}=0,35(mol)\\ a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2(1)\\ Fe+2HCl\to FeCl_2+H_2(2)\\ b,\text{Từ 2 PT: }1,5x+y=0,35(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow m_{Al}=0,1.27=2,7(g)\\ m_{Fe}=0,2.56=11,2(g)\)
\(c,n_{HCl(1)}=3n_{Al}=0,3(mol);n_{AlCl_3}=0,1(mol);n_{H_2(1)}=0,15(mol)\\ \Rightarrow m_{dd_{HCl(1)}}=\dfrac{0,3.36,5}{14,6\%}=75(g)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,1.133,5}{2,7+75-0,15.2}.100\%=17,25\%\)
\(n_{HCl(2)}=2n_{Fe}=0,4(mol);n_{FeCl_2}=n_{H_2(2)}=n_{Fe}=0,2(mol)\\ \Rightarrow m{dd_{HCl(2)}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%=22,92\%\)
a) 2Al + 6HCl --> 2AlCl3 + 3H2
Fe + 2HCl --> FeCl2 + H2
b) Gọi số mol Al, Fe lần lượt là a,b
=> 27a + 56b = 13,9
\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
2Al + 6HCl --> 2AlCl3 + 3H2
a----->3a--------->a------->1,5a______(mol)
Fe + 2HCl --> FeCl2 + H2
b------>2b-------->b----->b__________(mol)
=> 1,5a + b = 0,35
=> \(\left\{{}\begin{matrix}a=0,1=>m_{Al}=0,1.27=2,7\left(g\right)\\b=0,2=>m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)
c) nHCl = 3a + 2b = 0,7 (mol)
=> mHCl = 0,7.36,5 = 25,55(g)
=> \(m_{ddHCl}=\dfrac{25,55.100}{14,6}=175\left(g\right)\)
\(m_{dd\left(saupu\right)}=13,9+175-2.0,35=188,2\left(g\right)\)
\(\left\{{}\begin{matrix}m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\m_{FeCl_2}=0,2.127=25,4\left(g\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C\%\left(AlCl_3\right)=\dfrac{13,35}{188,2}.100\%=7,1\%\\C\%\left(FeCl_2\right)=\dfrac{25,4}{188,2}.100\%=13,5\%\end{matrix}\right.\)
thôi thì mình làm cho bn vậy, câu a ko làm dc đâu, làm câu b thôi, làm sao biết dc chất nào dư khi chỉ có số mol 1 chất?
nK2SO3=0.1367(mol)
mddH2SO4=Vdd.D=200.1,04=208(g)
K2SO3+H2SO4-->K2SO4+H2O+SO2
0.1367----0.1367----0.1367---------0.1367 (mol)
mddspu=100+208-0,1367.64=299.2512(g) ; mK2SO4=0,1367.174=23.7858(g)
==>C%=23.7858.100/299.512=7.94%
2)pt bn tự ghi nhé
ta có hệ pt: 56a+27b=11 và a+3b/2=8.96/22.4==>a=0.1, b=0.2
==>%Fe=0.1x56x100/11=50.9%
%Al=100%-50.9%=49.1%
b)nH2SO4= 0.7(mol)==>VddH2SO4=0.7/2=0.35(L)
\(n_{Al} = a\ ; n_{Fe} =b\\ \Rightarrow 27a + 56b = 11(1)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = 1,5a + b = \dfrac{8,96}{22,4} = 0,4(2)\\ (1)(2) \Rightarrow a = 0,2 ; b = 0,1\\ n_{HCl\ dư} = \dfrac{200.21,9\%}{36,5} - 0,2.3 - 0,1.2 = 0,4(mol)\\ m_{dd\ sau\ pư} = 11 + 200 - 0,4.2 = 210,2(gam)\\ C\%_{HCl} = \dfrac{0,4.36,5}{210,2}.100\% = 6,95\%\\ \)
\(C\%_{AlCl_3} = \dfrac{0,2.133,5}{210,2}.100\% = 12,7\%\\ C\%_{FeCl_2} = \dfrac{0,1.127}{210,2}.100\% = 6,04\%\)