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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: \(\left\{{}\begin{matrix}m_{hh}=56,8\left(g\right)\\m_{SO_2}=0,3\cdot64=19,2\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{SO_2}=237,6\left(g\right)\)
\(\Rightarrow C\%_{KCl}=\dfrac{59,6}{237,6}\cdot100\%\approx25,1\%\)
\(\text{Đặt }n_{Al}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 27x+56y=13,9(1)\\ n_{H_2}=\dfrac{7,84}{22,4}=0,35(mol)\\ a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2(1)\\ Fe+2HCl\to FeCl_2+H_2(2)\\ b,\text{Từ 2 PT: }1,5x+y=0,35(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow m_{Al}=0,1.27=2,7(g)\\ m_{Fe}=0,2.56=11,2(g)\)
\(c,n_{HCl(1)}=3n_{Al}=0,3(mol);n_{AlCl_3}=0,1(mol);n_{H_2(1)}=0,15(mol)\\ \Rightarrow m_{dd_{HCl(1)}}=\dfrac{0,3.36,5}{14,6\%}=75(g)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,1.133,5}{2,7+75-0,15.2}.100\%=17,25\%\)
\(n_{HCl(2)}=2n_{Fe}=0,4(mol);n_{FeCl_2}=n_{H_2(2)}=n_{Fe}=0,2(mol)\\ \Rightarrow m{dd_{HCl(2)}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%=22,92\%\)
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)
=> \(m_{ddHCl}=\dfrac{25,55.100}{14,6}=175\left(g\right)\)
\(m_{dd\left(saupu\right)}=13,9+175-2.0,35=188,2\left(g\right)\)
\(\left\{{}\begin{matrix}m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\m_{FeCl_2}=0,2.127=25,4\left(g\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C\%\left(AlCl_3\right)=\dfrac{13,35}{188,2}.100\%=7,1\%\\C\%\left(FeCl_2\right)=\dfrac{25,4}{188,2}.100\%=13,5\%\end{matrix}\right.\)
a) \(2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\)
\(Cu\left(OH\right)_2-^{t^o}\rightarrow CuO+H_2O\)
Gọi x,y lần lượt là số mol Fe(OH)3 và Cu(OH)2
=> \(\left\{{}\begin{matrix}107x+98y=20,5\\160.\dfrac{x}{2}+80y=16\end{matrix}\right.\)
=> x= 0,1 ; y=0,1
=> \(\%m_{Fe\left(OH\right)_3}=\dfrac{0,1.107}{20,5}.100=52,2\%\)
\(\%m_{Cu\left(OH\right)_2}=47,8\%\)
b) \(2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\)
\(Cu\left(OH\right)_2+H_2SO_4\rightarrow CuSO_4+2H_2O\)
\(n_{H_2SO_4}=0,1.\dfrac{3}{2}+0,1=0,25\left(mol\right)\)
\(m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)
\(m_{ddsaupu}=20,5+122,5=143\left(g\right)\)
\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,05.400}{143}.100=13,97\%\)
\(C\%_{CuSO_4}=\dfrac{0,1.160}{143}.100=11,19\%\)
c) \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(n_{Fe_2O_3}=0,05\left(mol\right);n_{CuO}=0,1\left(mol\right)\)
=> \(n_{H_2SO_4}=0,05.3+0,1=0,25\left(mol\right)\)
\(m_{ddH_2SO_4\left(pứ\right)}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)
=> \(m_{ddH_2SO_4\left(bđ\right)}=122,5.110\%=134,75\left(g\right)\)
nH2 = \(\dfrac{4,48}{22,4}\) = 0,2 (mol)
Zn + 2HCl ----> ZnCl2 + H2 (1)
ZnO + 2HCl ----> ZnCl2 + H2O (2)
nZn = nZnCl2 (1) = nH2 = 0,2 (mol)
=> mZn = 0.2 x 65 = 13 (g)
=> mZnO = 21,1 - 13 = 8,1 (g)
=> nZnO = 8,1/81 = 0.1 (mol)
nZnCl2 (2) = nZnO = 0.1 (mol)
C%ZnCl2 = \(\dfrac{152\left(0,2+0,1\right)}{21,1+200}\times100\%=20.62\%\)
PTHH:
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)
0,15 0,15
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
Ta có: \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=0,15\cdot100=15\left(g\right)\)
\(\Rightarrow m_{CaO}=20,6-15=5,6\left(g\right)\)
\(\Rightarrow\%m_{CaCO_3}=\dfrac{15\cdot100}{20,6}\approx73\%\)
\(\Rightarrow\%m_{CaO}=100\%-73\%=27\%\)
\(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,05 0,1 0,05 0,05
\(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
0,2 0,4 0,2
a) \(n_{Mg}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
\(m_{Mg}=0,05.24=1,2\left(g\right)\)
\(m_{MgO}=9,2-1,2=8\left(g\right)\)
0/0Mg = \(\dfrac{1,2.100}{9,2}=13,04\)0/0
0/0MgO = \(\dfrac{8.100}{9,2}=86,96\)0/0
b) Có : \(m_{MgO}=8\left(g\right)\)
\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
\(n_{HCl\left(tổng\right)}=0,1+0,4=0,5\left(mol\right)\)
\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)
\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)
c) \(n_{MgCl2\left(tổng\right)}=0,05+0,2=0,25\left(mol\right)\)
⇒ \(m_{MgCl2}=0,25.95=23,75\left(g\right)\)
\(m_{ddspu}=9,2+125-\left(0,05.2\right)=134,1\left(g\right)\)
\(C_{MgCl2}=\dfrac{23,75.100}{134,1}=17,71\)0/0
Chúc bạn học tốt
a)\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,05 0,1 0,05 0,05
PTHH: MgO + 2HCl → MgCl2 + H2O
Mol: 0,2 0,4 0,2
\(\%m_{Mg}=\dfrac{0,05.24.100\%}{9,2}=13,04\%;\%m_{MgO}=100-13,04=86,96\%\)
\(n_{MgO}=\dfrac{9,2-0,05.24}{40}=0,2\left(mol\right)\)
b,\(m_{HCl}=\left(0,1+0,4\right).36,5=18,25\left(g\right)\)
\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)
c,mdd sau pứ = 9,2+125-0,05.2 = 134,1 (g)
\(C\%_{ddMgCl_2}=\dfrac{\left(0,05+0,2\right).95.100\%}{134,1}=17,71\%\)
\(PTHH:Na_2SO_3+2HCl\rightarrow2NaCl+H_2O+SO_2\uparrow\\ K_2SO_3+2HCl\rightarrow2KCl+H_2O+SO_2\uparrow\\ n_{SO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_3}+n_{K_2SO_3}=0,2\\126n_{Na_2SO_3}+158n_{K_2SO_3}=28,4\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_3}=0,1\left(mol\right)\\n_{K_2SO_3}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\%_{m_{Na_2SO_3}}=\dfrac{0,1\cdot126}{28,4}\cdot100\%\approx44\%\\ \Rightarrow\%_{m_{K_2SO_3}}=100\%-44\%=56\%\)
\(n_{HCl}=0,1\cdot2+0,1\cdot2=0,4\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,4\cdot36,5=14,6\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{14,6}{200}\cdot100\%=7,3\%CC\)