\(S+O_2\underrightarrow{t^o}SO_2\) 
\(4FeS+7O_2\underrightarrow{t^o}2Fe_2O_3+4SO_2\)

\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ 2HCl+CuO\rightarrow CuCl_2+H_2O\\ Na_2O+2HCl\rightarrow2NaCl+H_2O\)

\(n_{Fe_2O_3}:a,n_{CuO}:b\) 
=> 160a + 80b = 32
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ pthh:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\) 
          a              3a 
           \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\) 
           b          b 
\(\left\{{}\begin{matrix}160a+80b=32\\3a+b=0,5\end{matrix}\right.\) 
=> a = 0,1 , b =0,2 
\(m_{Fe_2O_3}=0,1.160=16\left(g\right)\\ m_{CuO}=32-16=16\left(g\right)\) 
\(m_{H_2}=0,5.2=1\left(g\right)\)

gọi m_Fe  là a (mol) 
=> m_Cu là a (mol) 
=> m_Zn = 3a(mol) 
\(\%m_{Fe}=\dfrac{a}{a+a+3a}.100\%=20\%\\ \%m_{Cu}=\%m_{Fe}=20\%\\ \%m_{Zn}=100\%-20\%.2=60\%\) 
\(m_{Fe}=20.20\%=4\left(g\right)\\ m_{Cu}=m_{Fe}=4\left(g\right)\\ m_{Zn}=20-4.2=12\left(g\right)\)

\(pthh:S+O_2\underrightarrow{t^o}SO_2\left(1\right)\\ C+O_2\underrightarrow{t^o}CO_2\left(2\right)\\ m_S=\dfrac{40.20}{100}=8\left(g\right)\\ m_C=20-8=12\left(g\right)\\ n_S=\dfrac{8}{32}=0,25\left(mol\right)\\ n_C=\dfrac{12}{12}=1\left(mol\right)\\ n_{O_2}=n_S=0,25\left(mol\right)\\ n_{O_2\left(2\right)}=n_C=1\left(mol\right)\\ \Sigma n_{O_2}=1+0,25=1,25\left(mol\right)\\ V_{O_2}=1,25.22,4=28\left(l\right)\\ V_{kk}=\dfrac{28.100}{20}=140\left(l\right)\)

\(n_{NaOH}=0,1.1=0,1\left(mol\right)\\ pthh:NaOH+HCl\rightarrow NaCl+H_2O\)
           0,1            0,1         0,1         0,1 
\(C_{M\left(NaCl\right)}=\dfrac{0,1}{0,1+0,1}=0,5M\)

số mol của H₂ hình như sai bạn j đó ơi =))

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{73.20}{\dfrac{100}{36,5}}=0,4\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ LTL:\dfrac{0,1}{2}< \dfrac{0,4}{6}\) 
=> HCl dư 
\(m_{\text{dd}\left(saup\text{ư}\right)}=2,7+73-0,15.2=75,4\left(g\right)\\ C\%_{AlCl_3}=\dfrac{0,1.133,5}{75,4}.100\%=17,7\%\\ n_{HCl\left(d\text{ư}\right)}=0,4-0,3=0,1\left(mol\right)\\ C\%_{HCl\left(d\text{ư}\right)}=\dfrac{0,1.36,5}{75,4}.100\%=4,84\%\)

1
\(Al_2O_3\\ K_2SO_3\\ MgSO_4\\ BaCO_3\\ H_3PO_4\\ Ca\left(OH\right)_2\) 
\(ZnCl_2\\ NO_2\\ NaOH\\ Fe_2\left(SO_4\right)_3\\ AgCl\\ P_2O_5\\ Fe\left(NO_3\right)_2\\ AlCl_3\)
2
\(a.Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_{\text{ 4}}+2H_2O\\ b.CuO+2HCl\rightarrow CuCl_2+H_2O\\ c.Mg+\dfrac{1}{2}O_2\underrightarrow{t^o}MgO\\ d.ZnO+2HNO_3\rightarrow Zn\left(NO_3\right)_2+H_2O\\ 2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\\ Fe_xO_y+yH_2\underrightarrow{t^o}xFe+yH_2O\) 
3 
\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ pthh:4P+5O_2\underrightarrow{t^o}2P_2O_5\) 
           0,4     0,5      0,2 
\(m_P=0,4.31=12,4\left(g\right)\\ m_{P_2O_5}=0,2.142=28,4\left(g\right)\) 
4 
\(\%m_{Fe}=\dfrac{112.100}{160}=70\%\\ \%m_{O_2}=100\%-70\%=30\%\)

\(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\\ pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
          0,03    0,03                          0,03 
\(V_{H_2}=0,03.22,4=0,672\left(l\right)\\ m_{H_2SO_4}=0,03.98=2,94\left(g\right)\\ m_{\text{dd}H_2SO_4}=\dfrac{2,94.100}{19,6}=15\left(g\right)\)