Ta có: m_KL + m_O = m _oxit

⇒ m_O = 21,2 - 14 = 7,2 (g) \(\Rightarrow n_O=\dfrac{7,2}{16}=0,45\left(mol\right)\)

Có: \(2H^++O^{2-}\rightarrow H_2O\)

\(\Rightarrow n_{HCl}=n_{H^+}=2n_O=0,9\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{0,9}{0,1}=9\left(l\right)\)

a, \(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)

Ta có: \(n_{MnO_2}=\dfrac{34,8}{87}=0,4\left(mol\right)\)

Theo PT: \(n_{Cl_2}=n_{MnO_2}=0,4\left(mol\right)\Rightarrow V_{Cl_2}=0,4.22,4=8,96\left(l\right)\)

b, \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PT: \(Cl_2+H_2\underrightarrow{as}2HCl\)

Xét tỉ lệ: \(\dfrac{0,4}{1}< \dfrac{0,5}{1}\), ta được H₂ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{H_2\left(pư\right)}=n_{Cl_2}=0,4\left(mol\right)\\n_{HCl}=2n_{Cl_2}=0,8\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2\left(dư\right)}=0,5-0,4=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{HCl}=\dfrac{0,8.22,4}{\left(0,8+0,1\right).22,4}.100\%\approx88,89\%\\\%V_{H_2\left(dư\right)}\approx11,11\%\end{matrix}\right.\)

c, Ta có: \(m_{HCl}=0,8.36,5=29,2\left(g\right)\)

\(\Rightarrow C\%_{HCl}=\dfrac{29,2}{200}.100\%=14,6\%\)

d, \(n_{NaOH}=0,5.4=2\left(mol\right)\)

PT: \(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)

Xét tỉ lệ: \(\dfrac{0,4}{1}< \dfrac{2}{2}\), ta được NaOH dư.

Theo PT: \(\left\{{}\begin{matrix}n_{NaCl}=n_{NaClO}=n_{Cl_2}=0,4\left(mol\right)\\n_{NaOH\left(pư\right)}=2n_{Cl_2}=0,8\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{NaOH\left(dư\right)}=2-0,8=1,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{NaCl}}=C_{M_{NaClO}}=\dfrac{0,4}{0,5}=0,8\left(M\right)\\C_{M_{NaOH\left(dư\right)}}=\dfrac{1,2}{0,5}=2,4\left(M\right)\end{matrix}\right.\)

\(\left(K_2O,Al_2O_3,BaO\right)\underrightarrow{H_2O}\left(KOH,Ba\left(OH\right)_2\right),\left(Al_2O_3\right)\\ \left(Al_2O_3\right)\underrightarrow{dpnc}Al\\ \left(KOH,Ba\left(OH\right)_2\right)\underrightarrow{KHCO_3}\left(K_2CO_3,KHCO_3\right),\left(BaCO_3\right)\\ \left(BaCO_3\right)\underrightarrow{HCl}BaCl_2\underrightarrow{dp}Ba\\ \left(K_2CO_3,KHCO_3\right)\underrightarrow{HCl}\left(KCl\right)\underrightarrow{dp}K\\ K_2O+H_2O->2KOH\\ BaO+H_2O->Ba\left(OH\right)_2\\ Al_2O_3-dpnc->2Al+\dfrac{3}{2}O_2\\ KOH+KHCO_3->K_2CO_3+H_2O\\ Ba\left(OH\right)_2+2KHCO_3->BaCO_3+K_2CO_3+2H_2O\\ BaCO_3+2HCl->BaCl_2+H_2O+CO_2\\ BaCl_2-dp->Ba+Cl_2\\ K_2CO_3+2HCl->2KCl+H_2O+CO_2\\ KHCO_3+HCl->KCl+H_2O+CO_2\\ KCl-dp->K+\dfrac{1}{2}Cl_2\)

a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, Ta có: m _{dd tăng} = m_KL - m_H2

⇒ m_H2 = 16,6 - 15,6 = 1 (g) \(\Rightarrow n_{H_2}=\dfrac{1}{2}=0,5\left(mol\right)\)

Có: 27n_Al + 56n_Fe = 16,6 (1)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=0,5\left(2\right)\)

Từ (1) và (2) ⇒ n_Al = n_Fe = 0,2 (mol)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)

c, Theo PT: \(n_{HCl}=2n_{H_2}=1\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{1.36,5}{40\%}=91,25\left(g\right)\)

⇒ m _{dd sau pư} = 91,25 + 15,6 = 106,85 (g)

Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,2.133,5}{106,85}.100\%\approx24,98\%\\C\%_{FeCl_2}=\dfrac{0,2.127}{106,85}.100\%\approx23,77\%\end{matrix}\right.\)

\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{H_2}=\dfrac{16,6-15,6}{2}=0,5mol\\ n_{HCl}=1mol\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=16,6\\ 3a+2b=1\\ a=b=0,2\\ m_{Al}=0,2.27=5,4g\\ m_{Fe}=0,2.56=11,2g\\ m_{ddsau}=15,6+\dfrac{36,5}{0,4}=106,85g\\ C\%_{AlCl_3}=\dfrac{133,5.0,2}{106,85}.100\%=24,99\%\\ C\%_{FeCl_2}=\dfrac{127.0,2}{106,85}.100\%=23,77\%\)

\(n_{CuO}=n_{MgO}=n_{Fe_3O_4}=a\left(mol\right)\\ CuO+CO-t^{^0}->Cu+CO_2\\ Fe_3O_4+4CO-t^{^0}->3Fe+4CO_2\\ MgO+2HCl->MgCl_2+H_2O\\ Fe+2HCl->FeCl_2+H_2\\ 0,8=6a+2a\\ a=0,1\\ m=352a=35,2g\)

\(n_{tinh.thể}=a\left(g\right)\\ m_{CuSO_4\left(bđ\right)}=1877\cdot\dfrac{87,7}{187,7}=877g\\ C\%_{CuSO_4,12^{^0}C}=\dfrac{35,5}{135,5}=\dfrac{877-160\cdot\dfrac{a}{250}}{1877-a}\\ a=1019,133g\)

\(CaO+2HCl->CaCl_2+H_2O\\ a.n_{CaO}=\dfrac{5,6}{56}=0,1mol\\ m_{ddHCl}=\dfrac{0,2.36,5}{0,146}=50g\\ b.C\%_{CaCl_2}=\dfrac{111.0,1}{55,6}.100\%=19,96\%\)

Ta có: \(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)

\(n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\)

\(\Rightarrow\overline{M}_Y=\dfrac{3,2+8,8}{0,1+0,2}=40\left(g/mol\right)\)

\(\Rightarrow d_{Y/H_2}=\dfrac{\overline{M}_Y}{M_{H_2}}=\dfrac{40}{2}=20\)

\(SO_3+H_2O\rightarrow H_2SO_4\\ n_{H_2SO_4}=n_{SO_3}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ a,m=m_{H_2SO_4}=98.0,3=29,4\left(g\right)\\ b,C\%_{ddH_2SO_4}=\dfrac{29,4}{0,3.80+150}.100\approx16,897\%\\ c,H_2SO_4:Tính.axit\Rightarrow Quỳ.tím.hoá.đỏ\)