nhôm ko pứ với nước ở đk thường

\(n_{H_2SO_4}=0,6.1=0,6mol\\ n_{ZnO}=a;n_{Fe_2O_3}=b\\ a.ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ b.\Rightarrow\left\{{}\begin{matrix}81a+160b=40,3\\a+3b=0,6\end{matrix}\right.\\ \Rightarrow a=0,3;b=0,1\\ \%m_{ZnO}=\dfrac{0,3.81}{40,3}\cdot100\%=60,3\%\\ \%m_{Fe_2O_3}=100\%-60,3\%=39,7\%\\ c.n_{ZnSO_4}=n_{Zn}=0,3mol\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,1mol\\ C_{M_{ZnSO_4}}=\dfrac{0,3}{0,6}=0,5M\\ C_{M_{Fe_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,6}M\)

\(a.n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ n_{Mg}=a,n_{Na}=b\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ \Rightarrow\left\{{}\begin{matrix}24a+23b=9,4\\a+0,5b=0,3\end{matrix}\right.\\ \Rightarrow a=b=0,2\\ m_{Mg}=0,2.24=4,8g\\ m_{Na}=0,2.23=3,6g\\ b.n_{H_2SO_4}=0,5b+a=0,3mol\\ m_{ddH_2SO_4}=\dfrac{0,3.98}{10\%}\cdot100\%=294g\\ c.n_{MgSO_4}=n_{Mg}=0,2mol\\ n_{Na_2SO_4}=0,5n_{Na}=0,1mol\\ m_{MgSO_4}=120.0,2=24g\\ m_{Na_2SO_4}=142.0,1=14,2g\)

\(n_{CaCO_3}=\dfrac{50}{100}=0,5\left(mol\right)\\ a,PTHH:CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ b,n_{CO_2}=n_{CaCl_2}=n_{CaCO_3}=0,5\left(mol\right)\\ V_{CO_2\left(25^oC,1bar\right)}=0,5.24,79=12,395\left(l\right)\\ c,n_{HCl}=2.0,5=1\left(mol\right)\\ V_{ddHCl}=\dfrac{1}{0,2}=5\left(l\right)\\ d,m_{CaCl_2}=111.0,5=55,5\left(g\right)\)

\(n_{hh}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\\ C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ n_{C_2H_4}=n_{C_2H_4Br_2}=\dfrac{18,8}{188}=0,1\left(mol\right)\\ \Rightarrow\%V_{\dfrac{C_2H_4}{hh}}=\dfrac{0,1}{0,3}.100\%\approx33,333\%\Rightarrow\%V_{\dfrac{CH_4}{đktc}}\approx66,667\%\)

\(a.n_{BaCl_2}=\dfrac{20,8}{208}=0,1mol\\ n_{AgNO_3}=\dfrac{1,7}{170}=0,1mol\\ BaCl_2+2AgNO_3\rightarrow Ba\left(NO_3\right)_2+2AgCl\\ \Rightarrow\dfrac{0,1}{1}>\dfrac{0,1}{2}\Rightarrow BaCl_2.dư\\ BaCl_2+2AgNO_3\rightarrow Ba\left(NO_3\right)_2+2AgCl\)
0,05          0,1                   0,05                0,1
\(m_{Ba\left(NO_3\right)_2}=0,05.261=13,05g\\ m_{AgCl}=0,1.143,5=14,35g\\ m_{BaCl_2.dư}=\left(0,1-0,05\right).208=10,4g\\ b.V_{dd.sau}=0,03+0,07=0,1l\\ C_{M_{Ba\left(NO_3\right)_2}}=\dfrac{0,05}{0,1}=0,5M\\ C_{M_{BaCl_2.dư}}=\dfrac{0,1-0,05}{0,1}=0,5M\)

\(n_{Na_2CO_3.10H_2O}=\dfrac{85,8}{286}=0,3mol\\ m_{Na}=0,3.46=13,8g\\ m_C=0,3.12=3,6g\\ m_O=0,3.208=62,4g\\ m_H=0,3.20=6g\)