$\begin{array}{l} n_{SO_3}=\frac{1,6}{80}=0,02(mol)\\ \to T=\frac{n_{OH^-}}{n_{SO_3}}=\frac{0,2}{0,02}=10\\ \to \rm Tạo\,muối\,trung\,hòa,\,ion\,OH^-\,dư\,spu\\ SO_3+2NaOH\to Na_2SO_4+H_2O\\ Theo\,PT:\,n_{Na_2SO_4}=n_{SO_3}=0,02(mol)\\ \to C\%_{Na_2SO_4}=\frac{0,02.142}{200}.100\%=1,42\%\\ Theo\,PT:\,n_{NaOH(du)}=0,2-0,02.2=0,16(mol)\\ \to C\%_{NaOH(du)}=\frac{0,16.40}{200}.100\%=3,2\% \end{array}$

$\begin{array}{l} n_{Fe_2O_3}=\frac{108}{160}=0,675(mol)\\ Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\\ n_{FeCl_3}=2n_{Fe_2O_3}=1,35(mol)\\ m_{FeCl_3}=1,35.162,5=219,375(g) \end{array}$

$\begin{array}{l} n_{K_2O}=\frac{9,4}{94}=0,1(mol)\\ K_2O+H_2O\to 2KOH\\ n_{KOH}=2n_{K_2O}=0,2(mol)\\ m_{dd\,spu}=9,4+90,6=100(g)\\ \to C\%_{KOH}=\frac{0,2.56}{100}.100\%=11,2\% \end{array}$

$\begin{array}{l} a)\\ 2CH_3COOH+Na_2CO_3\to 2CH_3COONa+CO_2+H_2O\\ CH_3COOH+NaHCO_3\to CH_3COONa+CO_2+H_2O\\ b)\\ 40ml=0,04l;246,4=0,2464(l)\\ Dat\,n_{Na_2CO_3}=x(mol);n_{NaHCO_3}=y(mol)\\ \to 106x+84y=1,144(1)\\ n_{CO_2}=\frac{0,2464}{22,4}=0,011(mol)\\ Theo\,PT:\,x+y=0,011(2)\\ (1)(2)\to x=0,01(mol);y=0,001(mol)\\ Theo\,PT:\,\sum n_{CH_3COOH}=2x+y=0,021(mol)\\ \to C_{M\,CH_3COOH}=\frac{0,021}{0,04}=0,525M\\ c)\\ \rm Coi\,m_{dd\,CH_3COOH}\,không\,đáng\,kể\\ Theo\,PT:\,\sum n_{CH_3COONa}=\sum n_{CH_3COOH}=0,021(mol)\\ m_{dd\,spu}=0,021.60+25-0,011.44=25,776(g)\\ \to C\%_{CH_3COONa}=\frac{0,021.82}{25,776}.100\%\approx 6,68\% \end{array}$

$\begin{array}{l} a)Zn+2HCl\to ZnCl_2+H_2\\ b)\\ n_{Zn}=\frac{13}{65}=0,2(mol)\\ n_{H_2}=n_{Zn}=0,2(mol)\\ V_{H_2}=0,2.22,4=4,48(l)\\ c)\\ n_{CuO}=\frac{12}{80}=0,15(mol)\\ CuO+H_2\xrightarrow{t^o}Cu+H_2O\\ n_{CuO}<n_{H_2}\to H_2\rm\,dư\\ n_{H_2(pu)}=n_{CuO}=0,15(mol)\\ n_{H_2(du)}=0,2-0,15=0,05(mol)\\ m_{H_2(du)}=0,05.2=0,1(g) \end{array}$

$\begin{array}{l} 2)\\ a)\\ X:Al_2O_3\\ T:Al\\ Y:Al(NO_3)_3\\ Q:Al(OH)_3\\ Z:Al_2(SO_4)_3\\ -----------------\\ 4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ 2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\\ 4Al(NO_3)_3\xrightarrow{t^o}2Al_2O_3+12NO_2+3O_2\\ Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ b)\\ Al_2O_3+2KOH\to 2KAlO_2+H_2O\\ 2Al+2KOH+2H_2O\to 2KAlO_2+3H_2\\ Al(OH)_3+KOH\to KAlO_2+2H_2O\\ Al_2(SO_4)_3+6KOH\to 2Al(OH)_3↓+3K_2SO_4\\ c)\\ n_{Al(OH)_3}=\frac{3,9}{78}=0,05(mol)\\ Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3↓+3Na_2SO_4(1)\\ Al(OH)_3+NaOH\to NaAlO_2+2H_2O(2)\\ TH_1:\,\rm Chỉ\,xảy\,ra\,(1)\\ Theo\,PT:\,n_{NaOH}=3n_{Al(OH)_3}=0,15(mol)\\ \to V=\frac{0,15}{1}=0,15(l)\\ TH_2:\,\rm Xảy\,ra\,(1)\,và\,(2)\\ Theo\,PT:\,n_{NaAlO_2}=2n_{Al_2(SO_4)_3}-n_{Al(OH)_3}=0,15(mol)\\ Theo\,PT:\,\sum n_{NaOH}=0,15+0,1.6=0,75(mol)\\ \to V=\frac{0,75}{1}=0,75(l)\\ \to V\in \left\{0,15;0,75\right\}\\ 3)\\ a)\\ n_{SO_2}=\frac{13,44}{22,4}=0,6(mol)\\ Dat\,hoa\,tri\,A\,la\,n\\ A^0\to A^n+ne\\ S^{+6}+2e\to S^{+4}\\ Bao\,toan\,e:\,n.n_A=2n_{SO_2}=1,2\\ \to \frac{10,8n}{M_A}=1,2\\ \to M_A=9n\\ \to \begin{cases} n=3\\ M_A=27 \end{cases}\to A:Al\\ \rm Vậy\,A\,là\,nhôm\,(Al)\\ b)\\ n_{Al}=\frac{10,8}{27}=0,4(mol)\\ X:Al_2(SO_4)_3\\ Bao\,toan\,Al:\,n_{Al_2(SO_4)_3}=\frac{1}{2}n_{Al}=0,2(mol)\\ \to m_{Al_2(SO_4)_3(T)}=0,2.342.80\%=54,72(g)\\ Dat\,T\,la\,Al_2(SO_4)_3.xH_2O\\ \to n_{H_2O(T)}=\frac{25,92}{18}=1,44\\ \to n_{Al_2(SO_4)_3}:n_{H_2O}=(0,2.80\%):1,44=1:9\\ \to T:Al_2(SO_4)_3.9H_2O\\ c)\\ \Delta m_{dd}=m_{SO_2}-m_{Al}=0,6.64-10,8=27,6(g)\\ 4)\\ a)\\ n_{H_2}=\frac{13,44}{22,4}=0,6(mol)\\ m_{dd\,tang}=m_R-m_{H_2}\\ \to m=m_R=9,6+0,6.2=10,8(g)\\ Dat\,hoa\,tri\,R\,la\,n\\ 2R+2nHCl\to 2RCl_n+nH_2\\ Theo\,PT:\,n_R=\frac{2n_{H_2}}{n}=\frac{1,2}{n}\\ \to M_R=9n\to n=3;M_R=27\\ \to R:Al\\ Theo\,PT:\,n_{HCl}=2n_{H_2}=1,2(mol)\\ \to V_{dd\,HCl}=\frac{1,2}{0,5}=2,4(l)\\ b)\\ Theo\,PT:\,n_{AlCl_3}=\frac{2}{3}n_{H_2}=0,4(mol)\\ \to n_{H_2O(T)}=\frac{24,15-0,4.133,5.25\%}{18}=0,6(mol)\\ \to n_{AlCl_3(T)}:n_{H_2O(T)}=(0,4.25\%):0,6=1:6\\ \to T:AlCl_3.6H_2O\\ 5)\\ 200ml=0,2l;100ml=0,1l\\ n_{Al_2(SO_4)_3}=0,1.1=0,1(mol)\\ n_{Al(OH)_3}=\frac{12,48}{78}=0,16(mol)\\ Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3+3Na_2SO_4\\ TH_1:\rm Chỉ\,tạo\,ktua\\ \to n_{NaOH}=3n_{Al(OH)_3}=0,48(mol)\\ \to C_{M\,NaOH}=\frac{0,48}{0,2}=2,4M\\ TH_2:\rm Kết tủa bị hòa tan\\ Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3+3Na_2SO_4\\ Al(OH)_3+NaOH\to NaAlO_2+H_2O\\ Bao\,toan\,Al:\,n_{NaAlO_2}=2n_{Al_2(SO_4)_3}-n_{Al(OH)_3}=0,04(mol)\\ Theo\,PT:\,n_{NaOH}=3n_{Al(OH)_3}+4n_{NaAlO_2}=0,64(mol)\\ \to C_{M\,NaOH}=\frac{0,64}{0,2}=3,2M \end{array}$

#DominicHD247