Rainbow

\(2NaBr+Cl_2\rightarrow2NaCl+Br_2\left(1\right)\\ m_{giảm}=m_{Br_2}-m_{Cl_2}\\ \Leftrightarrow n_{NaCl\left(1\right)}=n_{NaBr\left(1\right)}=\dfrac{13,35}{160-71}=0,15\left(mol\right)\\ \Rightarrow\%m_{NaBr}=\dfrac{103.0,15}{42,6}.100\approx36,268\%\\ \Rightarrow\%m_{NaCl}\approx63,732\%\)

Bổ sung:

\(C\%_{ddNaBr\left(trongA\right)}=\dfrac{0,15.103}{200}.100=7,725\%\\ C\%_{ddNaCl\left(trongA\right)}=\dfrac{42,6-0,15.103}{200}.100=13,575\%\)

a) \(2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\)

\(Cu\left(OH\right)_2-^{t^o}\rightarrow CuO+H_2O\)

Gọi x,y lần lượt là số mol Fe(OH)3 và Cu(OH)2 

=> \(\left\{{}\begin{matrix}107x+98y=20,5\\160.\dfrac{x}{2}+80y=16\end{matrix}\right.\)

=> x= 0,1 ; y=0,1

=> \(\%m_{Fe\left(OH\right)_3}=\dfrac{0,1.107}{20,5}.100=52,2\%\)

\(\%m_{Cu\left(OH\right)_2}=47,8\%\)

b) \(2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\)

\(Cu\left(OH\right)_2+H_2SO_4\rightarrow CuSO_4+2H_2O\)

\(n_{H_2SO_4}=0,1.\dfrac{3}{2}+0,1=0,25\left(mol\right)\)

\(m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)

\(m_{ddsaupu}=20,5+122,5=143\left(g\right)\)

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,05.400}{143}.100=13,97\%\)

\(C\%_{CuSO_4}=\dfrac{0,1.160}{143}.100=11,19\%\)

c) \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(n_{Fe_2O_3}=0,05\left(mol\right);n_{CuO}=0,1\left(mol\right)\)

=> \(n_{H_2SO_4}=0,05.3+0,1=0,25\left(mol\right)\)

\(m_{ddH_2SO_4\left(pứ\right)}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)

=> \(m_{ddH_2SO_4\left(bđ\right)}=122,5.110\%=134,75\left(g\right)\)

PTHH: \(K_2O+2HCl\rightarrow2KCl+H_2O\)

            \(K_2SO_3+2HCl\rightarrow2KCl+SO_2\uparrow+H_2O\)

a) Ta có: \(\left\{{}\begin{matrix}n_{HCl}=\dfrac{200\cdot14,6\%}{36,5}=0,8\left(mol\right)\\n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_3}=0,3\left(mol\right)\\n_{K_2O}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{K_2O}=\dfrac{0,1\cdot94}{0,1\cdot94+0,3\cdot158}\cdot100\%\approx16,55\%\\\%m_{K_2SO_3}=83,45\%\end{matrix}\right.\)

b) Theo các PTHH: \(n_{KCl}=0,8\left(mol\right)\) \(\Rightarrow m_{KCl}=74,5\cdot0,8=59,6\left(g\right)\)

Mặt khác: \(\left\{{}\begin{matrix}m_{hh}=56,8\left(g\right)\\m_{SO_2}=0,3\cdot64=19,2\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{SO_2}=237,6\left(g\right)\)

\(\Rightarrow C\%_{KCl}=\dfrac{59,6}{237,6}\cdot100\%\approx25,1\%\)

\(n_{H_2}=\dfrac{7,84}{22,4}=0,35mol\)

Ba+2H₂O\(\rightarrow\)Ba(OH)₂+H₂ (1)

2Na+2H₂O\(\rightarrow\)2NaOH+H₂ (2)

- Gọi số mol Ba là x, số mol Na là y, ta có hệ phương trình:

\(\left\{{}\begin{matrix}137x+23y=34,3\\x+\dfrac{1}{2}y=0,35\end{matrix}\right.\)

Giải ra x=0,2 và y=0,3

\(\%Ba=\dfrac{0,2.137.100}{34,3}\approx80\%\)

%Na\(\approx\)20%

- Dung dịch A gồm: Ba(OH)₂ 0,2 mol và NaOH 0,3mol

\(Ba\left(OH\right)_2+K_2SO_4\rightarrow BaSO_{4_{ }}+2KOH\)(3)

\(n_{BaSO_4}=n_{Ba\left(OH\right)_2}=0,1mol\)

\(m_{BaSO_4}=0,1.233=23,3g\)

- Sau phản ứng (3) có:0,15mol NaOH và 0,2 mol KOH

\(C_{M_{NaOH}}=\dfrac{n}{v}=\dfrac{0,15}{0,5}=0,3M\)

\(C_{M_{KOH}}=\dfrac{n}{v}=\dfrac{0,1}{0,5}=0,2M\)

\(Ba\left(OH\right)_2+CuCl_2\rightarrow BaCl_2+Cu\left(OH\right)_2\)(4)

\(2NaOH+CuCl_2\rightarrow2NaCl+Cu\left(OH\right)_2\)(5)

- Theo PTHH (4) và (5) ta có:

\(n_{CuCl_2}=n_{Ba\left(OH\right)_2}+\dfrac{1}{2}n_{NaOH}=0,1+\dfrac{1}{2}.0,15=0,175mol\)\(m_{dd_{CuCl_2}}=\dfrac{0,175.135.100}{10}=236,25g\)

- Dung dịch C có: BaCl₂ 0,1 mol và NaCl 0,15 mol

\(m_{dd}=\dfrac{1}{2}m_{dd_A}+m_{dd_{CuCl_2}}-m_{Cu\left(OH\right)_2}\)

\(m_{dd}=\dfrac{1}{2}.500.1,2+236,25-0,175.98=519,1g\)

\(C\%_{BaCl_2}=\dfrac{0,1.208.100}{519,1}\approx4\%\)

\(C\%_{NaCl}=\dfrac{0,15.58,5.100}{519,1}\approx1,7\%\)

Gọi : \(\left\{{}\begin{matrix}n_{Al_2O_3}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\)⇒ 102a + 65b = 2,505(1)

\(Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O\\ Zn + 2HCl \to ZnCl_2 + H_2\)

Muối gồm : \(\left\{{}\begin{matrix}AlCl_3:2a\left(mol\right)\\ZnCl_2:b\left(mol\right)\end{matrix}\right.\)⇒ 133,5.2a + 136b = 6,045(2)

Từ (1)(2) suy ra : a = 0,015 ; b = 0,015

Vậy :

\(\%m_{Al_2O_3} = \dfrac{0,015.102}{2,505}.100\% = 61,08\%\\ \%m_{Zn} = 100\% - 61,08\% = 38,92\%\)

Theo PTHH : \(n_{HCl} = 6a + 2b = 0,12(mol)\\ \Rightarrow C\%_{HCl} = \dfrac{0,12.36,5}{200}.100\% = 2,19\%\)

\(n_{NaOH}=1,5.0,15=0,225mol\)

Gọi \(\left\{{}\begin{matrix}n_{CH_3COOH}=x\\n_{CH_3COOC_2H_5}=y\end{matrix}\right.\) ( mol )

\(\rightarrow60x+88y=15,6\left(g\right)\) (1)

\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

         x                    x                  x                       ( mol )

\(CH_3COOC_2H_5+NaOH\rightarrow CH_3COONa+C_2H_5OH\)

        y                          y                  y                              ( mol )

\(n_{NaOH}=x+y=0,225\left(mol\right)\) (2)

\(\left(1\right);\left(2\right)\rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,075\end{matrix}\right.\) ( mol )

\(\rightarrow\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,15.60}{15,6}.100\%=57,69\%\\\%m_{CH_3COOC_2H_5}=100\%-57,69\%=42,31\%\end{matrix}\right.\)

\(m_{CH_3COONa}=\left(0,15+0,075\right).82=18,45g\)

a)Gọi x,y lần lượt là số mol của Al, Fe trong hỗn hợp ban đầu (x,y>0)

Sau phản ứng hỗn hợp muối khan gồm: \(\left\{{}\begin{matrix}AlCl_3:x\left(mol\right)\\FeCl_2:y\left(mol\right)\end{matrix}\right.\)

Ta có hệ phương trình: \(\left\{{}\begin{matrix}27x+56y=13,9\\133,5x+127y=38\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\approx0,0896\\y\approx0,205\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,0896\cdot27\cdot100\%}{13,9}\approx17,4\%\\\%m_{Fe}=\dfrac{0,205\cdot56\cdot100\%}{13,9}\approx82,6\%\end{matrix}\right.\)

Theo Bảo toàn nguyên tố Cl, H ta có:\(n_{H_2}=\dfrac{n_{HCl}}{2}=\dfrac{3n_{AlCl_3}+2n_{FeCl_2}}{2}\\ =\dfrac{3\cdot0,0896+2\cdot0,205}{2}=0,3394mol\\ \Rightarrow V_{H_2}=0,3394\cdot22,4\approx7,6l\)