\(n_{HCl}=0,1.0,3=0,03\left(mol\right)\)

=> \(n_{H_2O}=\dfrac{0,03}{2}=0,015\left(mol\right)\)

Theo ĐLBTKL: \(m_{oxit}+m_{HCl}=m_{muối}+m_{H_2O}\)

=> m_muối = 3,425 + 0,03.36,5 - 0,015.18 = 4,25(g) 

\(n_{HCl}=1,4.0,5=0,7\left(mol\right)\\ n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ Vì:\dfrac{0,7}{2}>\dfrac{0,2}{1}\Rightarrow HCldư\\ \Rightarrow ddsau:\left\{{}\begin{matrix}HCl\left(dư\right)\\CuCl_2\end{matrix}\right.\\ V_{ddsau}=V_{ddHCl}=500\left(ml\right)=0,5\left(l\right)\\ n_{CuCl_2}=n_{CuO}=0,2\left(mol\right)\\ n_{HCl\left(dư\right)}=0,7-0,2.2=0,3\left(mol\right)\\ m_{CuCl_2}=135.0,2=27\left(g\right)\\ m_{HCl\left(dư\right)}=0,3.36,5=10,95\left(g\right)\\ C_{MddHCl}=\dfrac{0,3}{0,5}=0,6\left(M\right)\\ C_{MddCuCl_2}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)

a) 

n_NaOH = 0,04.1 = 0,04 (mol)

PTHH: NaOH + HCl --> NaCl + H₂O

           0,04--->0,04

=> n_{HCl(pư với X)} = 0,2.1 - 0,04 = 0,16 (mol)

Gọi số mol CuO, Fe₂O₃ là a, b (mol)

=> 80a + 160b = 4,8 (1)

PTHH: CuO + 2HCl --> CuCl₂ + H₂O

              a----->2a

            Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

               b----->6b

=> 2a + 6b = 0,16 (2)

(1)(2) => a = 0,02; b = 0,02

=> \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,02.80}{4,8}.100\%=33,33\%\\\%m_{Fe_2O_3}=\dfrac{0,02.160}{4,8}.100\%=66,67\%\end{matrix}\right.\)

b) Chất rắn thu được gồm CuO, Fe₂O₃

Bảo toàn Cu: n_CuO = 0,02 (mol)

Bảo toàn Fe: n_Fe2O3 = 0,02 (mol)

=> m = 0,02.80 + 0,02.160 = 4,8 (g)

a) Gọi số mol Zn, Fe là a, b (mol)

=> 65a + 56b = 7,35 (1)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           a---->2a------->a------>a

             Fe + 2HCl --> FeCl₂ + H₂

             b------>2b----->b------>b

=> \(a+b=\dfrac{2,688}{22,4}=0,12\left(mol\right)\)

=> a + b = 0,12 (2)

(1)(2) => a = 0,07; b = 0,05

=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,07.65}{7,35}.100\%=61,9\%\\\%m_{Fe}=\dfrac{0,05.56}{7,35}.100\%=38,1\%\end{matrix}\right.\)

b) n_HCl(dư) = 0,3.1 - 0,07.2 - 0,05.2 = 0,06 (mol)

PTHH: Ca(OH)₂ + 2HCl --> CaCl₂ + 2H₂O

             0,03<-----0,06

=> \(x=C_{M\left(ddCa\left(OH\right)_2\right)}=\dfrac{0,03}{0,1}=0,3M\)

c) Chất rắn thu được là Fe₂O₃

Bảo toàn Fe: \(n_{Fe_2O_3}=0,025\left(mol\right)\)

=> \(a=m_{Fe_2O_3}=0,025.160=4\left(g\right)\)

Kết tủa thu được là Fe(OH)₂

Bảo toàn Fe: \(n_{Fe\left(OH\right)_2}=0,05\left(mol\right)\)

=> \(m=m_{Fe\left(OH\right)_2}=0,05.90=4,5\left(g\right)\)

\(n_{SO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right);n_{NaOH}=1.0,15=0,15\left(mol\right)\\ Vì:1< \dfrac{n_{NaOH}}{n_{SO_2}}=\dfrac{0,15}{0,1}=1,5< 2\\ \Rightarrow Sp:Na_2SO_3,NaHSO_3\\ PTHH:2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\left(1\right)\\ NaOH+SO_2\rightarrow NaHSO_3\left(2\right)\\ Đặt:n_{NaOH\left(1\right)}=a\left(mol\right);n_{NaOH\left(2\right)}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}a+b=0,15\\0,5a+b=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow m_{muối}=0,5.a.126+b.104=11,5\left(g\right)\)

Tham khảo:

Chúc bn học tốt

\(a,n_{H_2}=\dfrac{2,576}{22,4}=0,115\left(mol\right)\\ Đặt:n_{Mg}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}95a+133,5b=10,475\\a+1,5b=0,115\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\\b=0,05\end{matrix}\right.\\ \%m_{Mg}=\dfrac{0,04.24}{0,04.24+0,05.27}.100\approx41,558\%\Rightarrow\%m_{Al}\approx58,442\%\\ b,n_{HCl}=2.n_{H_2}=2.0,115=0,23\left(mol\right)\\ \Rightarrow x=C\%_{ddHCl}=\dfrac{0,23.36,5}{100}.100=8,395\%\)

a) \(n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

          0,05<-----------0,05---->0,075

=> \(\%Al=\dfrac{0,05.27}{14,15}.100\%=9,54\%\)

=> \(\%Cu=\dfrac{14,15-0,05.27}{14,15}.100\%=90,46\%\)

b) \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)

c) \(n_{Cu}=\dfrac{14,15-0,05.27}{64}=0,2\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

          0,05->0,0375

           2Cu + O₂ --t^o--> 2CuO 

            0,2-->0,1

=> \(V_{O_2}=\left(0,1+0,0375\right).22,4=3,08\left(l\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ m_{AlCl_3}=6,675\left(mol\right)\\ n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,05\left(mol\right)\\ \Rightarrow m_A=0,05.27=1,35\left(g\right);m_{Cu}=14,15-1,35=12,8\left(g\right)\\ \%m_{Cu}=\dfrac{12,8}{14,15}.100\approx90,459\%\\ \Rightarrow\%m_{Al}\approx9,541\%\\ b,n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ \Rightarrow V=V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{O_2}=\dfrac{3}{4}.n_{Al}+\dfrac{1}{2}.n_{Cu}=\dfrac{3}{4}.0,05+\dfrac{1}{2}.0,2=0,0875\left(mol\right)\)

\(\Rightarrow V_{O_2\left(đktc\right)}=0,0875.22,4=1,96\left(l\right)\)

\(n_{AgNO_3}=0,009.2=0,018\left(mol\right)\\ n_{NaX}=1,5.0,01=0,015\left(mol\right)\\ NaX+AgNO_3\rightarrow AgX\downarrow+NaNO_3\\ Vì:\dfrac{0,018}{1}>\dfrac{0,015}{1}\\ \Rightarrow AgNO_3dư\\ \Rightarrow n_{AgX}=n_{NaX}=0,015\left(mol\right)\\ \Rightarrow M_{AgX}=\dfrac{2,82}{0,015}=188\left(\dfrac{g}{mol}\right)=M_{Ag}+M_X=M_X+108\\ \Leftrightarrow M_X=80\left(\dfrac{g}{mol}\right)\\ \Rightarrow X:Brom\left(Br=80\right)\\ 2,dd.sau.phản.ứng:AgNO_3\left(dư\right),NaNO_3\\ n_{AgNO_3\left(p.ứ\right)}=n_{NaNO_3}=n_{NaBr}=0,015\left(mol\right)\\ \Rightarrow n_{AgNO_3\left(dư\right)}=0,018-0,015=0,003\left(mol\right)\)

\(n_{Cl_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ Cl_2+2NaI\rightarrow2NaCl+I_2\\ n_{NaI}=0,2.0,6=0,12\left(mol\right)\\ Vì:\dfrac{0,12}{2}>\dfrac{0,05}{1}\\ \Rightarrow NaIdư\\ ddX:NaI,NaCl\\ NaI_{dư}+AgNO_3\rightarrow AgI\downarrow\left(vàng.đậm\right)+NaNO_3\\ NaCl+AgNO_3\rightarrow AgCl\downarrow\left(trắng\right)+NaNO_3\\ n_{NaI\left(p.ứ\right)}=n_{NaCl}=2.n_{Cl_2}=0,1\left(mol\right)\\ n_{NaI\left(dư\right)}=0,12-0,1=0,02\left(mol\right)\\ n_{AgCl}=n_{NaCl}+n_{NaI\left(dư\right)}=0,1+0,02=0,12\left(mol\right)\\ \Rightarrow m=m_{\downarrow}=m_{AgCl}=143,5.0,12=17,22\left(g\right)\)

\(2M+3Cl_2\rightarrow2MCl_3\\ n_{Cl_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\\ \Rightarrow n_M=\dfrac{2.0,075}{3}=0,05\left(mol\right)\\ M+2HCl\rightarrow MCl_2+H_2\\ n_{HCl}=2n_M=2.0,05=0,1\left(mol\right)\\ \Rightarrow V=V_{ddHCl}=\dfrac{0,1}{2}=0,05\left(l\right)=50\left(ml\right)\)