Câu 2:
\(n_{CO_2}=\dfrac{3.36}{22,4}=0,15(mol)\\ a,PTHH:Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2\uparrow\\ \Rightarrow n_{HCl}=2n_{CO_2}=0,3(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,3}{2}=0,15(l)\\ d,n_{Na_2CO_3}=n_{CO_2}=0,15(mol)\\ \Rightarrow m_{Na_2CO_3}=0,15.106=15,9(g)\\ \Rightarrow \%_{Na_2CO_3}=\dfrac{15,9}{20}.100\%=79,5\%\\ \Rightarrow \%_{NaCl}=100\%-79,5\%=20,5\%\\ c,n_{NaCl}=n_{HCl}=0,3(mol)\\ \Rightarrow m_{NaCl}=0,3.58,5=17,55(g)\)
Câu 1:
\(a,(1)Cu(OH)_2\xrightarrow{t^o}CuO+H_2O\\ (2)CuO+2HCl\to CuCl_2+H_2O\\ (3)CuCl_2+2AgNO_3\to Cu(NO_3)_2+2AgCl\downarrow\\ (4)Cu(NO_3)_2+Fe\to Fe(NO_3)_2+Cu\\ (5)Fe(NO_3)_2+2NaOH\to Fe(OH)_2\downarrow+2NaNO_3\)
\(b,(1)Fe(NO_3)_3+3NaOH\to Fe(OH)_3\downarrow+3NaNO_3\\ (2)2Fe(OH)_3\xrightarrow{t^o}Fe_2O_3+3H_2O\\ (3)Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2\\ (4)Fe+2HCl\to FeCl_2+H_2\)
\(c,(1)2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\\ (2)Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ (3)Al_2(SO_4)_3+3BaCl_2\to 2AlCl_3+3BaSO_4\downarrow\\ (4)AlCl_3+3AgNO_3\to Al(NO_3)_3+3AgCl\downarrow\\ (5)2Al(NO_3)_3+3Mg\to 3Mg(NO_3)_2+2Al\)
Câu 3:
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=0,2(mol)\\ \Rightarrow \%_{Mg}=\dfrac{0,2.24}{8,8}.100\%=54,55\%\\ \Rightarrow \%_{MgO}=100\%-54,55\%=45,45\%\\ b,n_{MgO}=\dfrac{8,8-0,2.2,4}{40}=0,1(mol)\\ \Rightarrow \Sigma n_{HCl}=0,2.2+0,1.2=0,6(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,3}=2M\)