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\(\text{Đặt }n_{Al}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 27x+56y=13,9(1)\\ n_{H_2}=\dfrac{7,84}{22,4}=0,35(mol)\\ a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2(1)\\ Fe+2HCl\to FeCl_2+H_2(2)\\ b,\text{Từ 2 PT: }1,5x+y=0,35(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow m_{Al}=0,1.27=2,7(g)\\ m_{Fe}=0,2.56=11,2(g)\)
\(c,n_{HCl(1)}=3n_{Al}=0,3(mol);n_{AlCl_3}=0,1(mol);n_{H_2(1)}=0,15(mol)\\ \Rightarrow m_{dd_{HCl(1)}}=\dfrac{0,3.36,5}{14,6\%}=75(g)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,1.133,5}{2,7+75-0,15.2}.100\%=17,25\%\)
\(n_{HCl(2)}=2n_{Fe}=0,4(mol);n_{FeCl_2}=n_{H_2(2)}=n_{Fe}=0,2(mol)\\ \Rightarrow m{dd_{HCl(2)}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%=22,92\%\)
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)
=> \(m_{ddHCl}=\dfrac{25,55.100}{14,6}=175\left(g\right)\)
\(m_{dd\left(saupu\right)}=13,9+175-2.0,35=188,2\left(g\right)\)
\(\left\{{}\begin{matrix}m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\m_{FeCl_2}=0,2.127=25,4\left(g\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C\%\left(AlCl_3\right)=\dfrac{13,35}{188,2}.100\%=7,1\%\\C\%\left(FeCl_2\right)=\dfrac{25,4}{188,2}.100\%=13,5\%\end{matrix}\right.\)
CH3COOH + NaOH => CH3COONa + H2O
nNaOH = CM.V = 0.15 (mol)
Theo pt ==> nCH3COOH = 0.15 (mol)
==> mCH3COOH = n.M = 0.15x60 = 9 (g)
==> mC2H5OH = mhh - mCH3COOH = 13.6 - 9 = 4.6 (g)
%mC2H5OH = 4.6x100/13.6 = 33.82 (%)
%mCH3COOH = 100% - 33.82 = 66.18 (%)
Ta có: nCH3COONa = 0.15 (mol)
==> mCH3COONa = 0.15x82x90/100 = 11.07 (g)
PTHH: \(K_2O+2HCl\rightarrow2KCl+H_2O\)
\(K_2SO_3+2HCl\rightarrow2KCl+SO_2\uparrow+H_2O\)
a) Ta có: \(\left\{{}\begin{matrix}n_{HCl}=\dfrac{200\cdot14,6\%}{36,5}=0,8\left(mol\right)\\n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_3}=0,3\left(mol\right)\\n_{K_2O}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{K_2O}=\dfrac{0,1\cdot94}{0,1\cdot94+0,3\cdot158}\cdot100\%\approx16,55\%\\\%m_{K_2SO_3}=83,45\%\end{matrix}\right.\)
b) Theo các PTHH: \(n_{KCl}=0,8\left(mol\right)\) \(\Rightarrow m_{KCl}=74,5\cdot0,8=59,6\left(g\right)\)
Mặt khác: \(\left\{{}\begin{matrix}m_{hh}=56,8\left(g\right)\\m_{SO_2}=0,3\cdot64=19,2\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{SO_2}=237,6\left(g\right)\)
\(\Rightarrow C\%_{KCl}=\dfrac{59,6}{237,6}\cdot100\%\approx25,1\%\)
a)
Gọi $n_{ZnCl_2} = a(mol) ; n_{FeCl_3} = b(mol)$
Ta có :
$m_{hh} = 136a + 162,5b = 29,85(gam)$
$n_{NaOH} = 2a + 3b = 0,5(mol)$
Suy ra a = b = 0,1(mol)$
$m_{ZnCl_2} = 0,1.136 = 13,6(gam)$
$m_{FeCl_3} = 0,1.162,5 = 16,25(gam)$
b)
$m_{Zn(OH)_2} = 0,1.99 = 9,9(gam)$
$m_{Fe(OH)_3} = 0,1.107 = 10,7(gam)$
Sau phản ứng :
$m_{dd} = 29,85 + 500.1,1 - 9,9 - 10,7 = 559,25(gam)$
$n_{NaCl} = n_{NaOH} = 0,5(mol)$
$C\%_{NaCl} = \dfrac{0,5.58,5}{559,25}.100\% = 5,23\%$
c)
$V_{dd} = 0,2+ 0,5 = 0,7(lít)$
$[Na^+] = \dfrac{0,5}{0,7} = 0,714M$
$[Cl^-] = \dfrac{0,5}{0,7} = 0,714M$
(Thiếu nồng độ $H_2SO_4$)
\(n_{NaOH}=1,5.0,15=0,225mol\)
Gọi \(\left\{{}\begin{matrix}n_{CH_3COOH}=x\\n_{CH_3COOC_2H_5}=y\end{matrix}\right.\) ( mol )
\(\rightarrow60x+88y=15,6\left(g\right)\) (1)
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
x x x ( mol )
\(CH_3COOC_2H_5+NaOH\rightarrow CH_3COONa+C_2H_5OH\)
y y y ( mol )
\(n_{NaOH}=x+y=0,225\left(mol\right)\) (2)
\(\left(1\right);\left(2\right)\rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,075\end{matrix}\right.\) ( mol )
\(\rightarrow\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,15.60}{15,6}.100\%=57,69\%\\\%m_{CH_3COOC_2H_5}=100\%-57,69\%=42,31\%\end{matrix}\right.\)
\(m_{CH_3COONa}=\left(0,15+0,075\right).82=18,45g\)