Bài 2.

a) \(\left(1\right)2HCl\rightarrow H_2+Cl_2\)

    \(\left(2\right)Cl_2+H_2\rightarrow2HCl\)

    \(\left(3\right)2Fe+3Cl_2\underrightarrow{t_o}2FeCl_3\)

    \(\left(4\right)2FeCl_3+Fe\rightarrow3FeCl_2\)

    \(\left(5\right)2NaOH+Cl_2\rightarrow H_2O+NaCl+NaClO\)

b) \(\left(1\right)2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(\left(2\right)Cl_2+H_2\rightarrow2HCl\)

\(\left(3\right)2HCl+FeO\rightarrow FeCl_2+H_2O\)

\(\left(4\right)3FeCl_2+4HNO_3\rightarrow2FeCl_3+Fe\left(NO_3\right)_3+NO+2H_2O\)

\(\left(5\right)2FeCl_3+Fe\rightarrow3FeCl_2\)

\(\left(6\right)FeCl_3+3AgNO_3\rightarrow Fe\left(NO_3\right)_3+3AgCl\downarrow\)

Câu 1:

a)

\(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

\(Mg+Cl_2\underrightarrow{t^o}MgCl_2\)

\(2NaBr+Cl_2\rightarrow2NaCl+Br_2\)

\(2KI+Cl_2\rightarrow2KCl+I_2\)

\(H_2+Cl_2\underrightarrow{as}2HCl\)

\(2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O\)

\(Cl_2+H_2O⇌HCl+HClO\)

b) 

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2\uparrow+H_2O\)

\(HCl+AgNO_3\rightarrow AgCl\downarrow+HNO_3\)

\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

BT e: \(3n_{Al}+2n_{Mg}=4n_{O_2}+2n_{Cl_2}\)

\(\Rightarrow3\cdot0,1+2\cdot0,05=4\cdot0,05+2\cdot x\)

\(\Rightarrow x=0,1mol\)

\(m=m_{Al}+m_{Mg}+m_{O_2}+m_{Cl_2}\)

    \(=0,1\cdot27+0,05\cdot24+0,05\cdot2\cdot16+0,1\cdot35,5\cdot2\)

    \(=12,6g\)

Theo bảo toàn electron ta có: \(3\cdot n_{Al}+2\cdot n_{Mg}=2\cdot n_{Cl_2}+4\cdot N_{O_2}\)

\(\Rightarrow3\cdot0,1+2\cdot0,05=4\cdot0,05+2x\Rightarrow x=0,2\)

\(\Rightarrow m_Z=m_X+m_Y=0,1\cdot27+0,05\cdot24+0,05\cdot32+0,2\cdot71=19,7g\)

Gọi khối lượng NaCl trong 1900 gam dd NaCl bão hòa ở 90^oC là a (gam)

Có: \(S_{NaCl\left(90^oC\right)}=\dfrac{a}{1900-a}.100=50\)

=> a = \(\dfrac{1900}{3}\left(g\right)\)

=> \(m_{H_2O}=1900-\dfrac{1900}{3}=\dfrac{3800}{3}\left(g\right)\)

Gọi khối lượng NaCl trong dd NaCl bão hòa ở 0^oC là b (gam)

Có: \(S_{NaCl\left(0^oC\right)}=\dfrac{b}{\dfrac{3800}{3}}.100=35\)

=> \(b=\dfrac{1330}{3}\left(g\right)\)

=> m_{NaCl (tách ra)} = \(\dfrac{1900}{3}-\dfrac{1330}{3}=190\left(g\right)\)

\(m_{NaCl\left(tách\right)}=m_{NaCl\left(dd.bão.hoà.90^oC\right)}-m_{NaCl\left(dd.bão.hoà.0^oC\right)}\\ =\dfrac{1900}{100}.50-\dfrac{1900}{100}.35=285\left(g\right)\)

Em xem có gì không hiểu thì hỏi lại nha!

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\\ \Rightarrow m_{FeCl_2}=162,5.0,2=32,5\left(g\right)\\ \Rightarrow m=32,5\left(g\right)\)

Đề có cho HCl đâu em

a: \(Cu+Cl_2\rightarrow CuCl_2\)(ĐK: t độ)

\(2Fe+3Cl_2\rightarrow2FeCl_3\)(ĐK: t độ)

\(2Na+Br_2\rightarrow2NaBr\)

\(2Al+3Br_2\rightarrow2AlBr_3\)

b: \(H_2+Cl_2\rightarrow2HCl\)(ĐK: ánh sáng)

\(H_2+Br_2\rightarrow2HBr\)

c: \(Cl_2+2KI\rightarrow I_2+2KCl\)

\(Cl_2+2KBr\rightarrow Br_2+2KCl\)

\(Br_2+2KI\rightarrow2KBr+I_2\)

a)

\(n_{hhA}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ Đặt:n_{Cl_2}=a\left(mol\right);n_{O_2}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow a+b=0,5\left(1\right)\\ Mà:m_{hhA}=m_{muối}-m_{hhB}=42,34-16,98=25,36\left(g\right)\left(2\right)\\ \left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}a+b=0,5\\71a+32b=25,36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,24\\b=0,26\end{matrix}\right.\\ \Rightarrow\%V_{Cl_2}=\dfrac{0,24}{0,5}.100=48\%\Rightarrow \%V_{O_2}=\dfrac{0,26}{0,5}.100=52\%\)

b) Đặt x,y là số mol của Mg, Al trong hhB (x,y>0) (mol)

=> 24x+27y=16,98(3)

Áp dụng ĐLBT electron:

\(n_{e.cho}=n_{e.nhận}\\ \Leftrightarrow2.n_{Mg}+3.n_{Al}=2.n_{Cl_2}+4.n_{O_2}\\ \Leftrightarrow2x+3y=2.0,24+4.0,26=1,52\left(4\right)\\ \left(3\right),\left(4\right)\Rightarrow\left\{{}\begin{matrix}2x+3y=1,52\\24x+27y=16,98\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,55\\y=0,14\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,55.24}{16,98}.100\approx77,739\%\\ \Rightarrow\%m_{Al}\approx22,261\%\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\n_{H_2}=\dfrac{2,16}{22,4}=0,09\left(mol\right)\\ \Rightarrow \left\{{}\begin{matrix}1,5a+b=0,09\\27a+56b=2,76\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\\b=0,03\end{matrix}\right.\\ \Rightarrow\%m_{Al}=\dfrac{0,04.27}{2,76}.100\approx39,13\%\\ \Rightarrow\%m_{Fe}\approx100\%-39,13\%\approx60,87\%\)

\(b,V_{ddsau}=V_{ddHCl}=0,2\left(l\right)\\ n_{AlCl_3}=n_{Al}=0,04\left(mol\right);n_{FeCl_2}=n_{Fe}=0,03\left(mol\right)\\ \Rightarrow C_{MddFeCl_2}=\dfrac{0,03}{0,2}=0,15\left(M\right)\\ C_{MddAlCl_3}=\dfrac{0,04}{0,2}=0,2\left(M\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

  \(x\)                                      \(1,5x\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

 \(y\)                                    \(y\)

Có \(27x+56y=2,76\left(1\right)\)

\(1,5x+y=\dfrac{2,016}{22,4}=0,09\left(2\right)\)

Từ (1) và (2)\(\Rightarrow\left\{{}\begin{matrix}x=0,04\\y=0,03\end{matrix}\right.\)

\(\%m_{Al}=\dfrac{0,04\cdot27}{2,76}\cdot100\%=39,13\%\)

\(\%m_{Fe}=100\%-39,13\%=60,87\%\)

\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)

\(x\)                                                          \(x\)

\(MgCO_2+2HCl\rightarrow MgCl_2+H_2O+CO_2\uparrow\)

\(y\)                                                             \(y\)

Thu đc kết tủa \(BaSO_4\)

\(n_{BaSO_4}=\dfrac{1,7475}{233}=0,0075mol=n_{Ba\left(OH\right)_2dư}\)

\(\Rightarrow n_{Ba\left(OH\right)_2PƯ}=0,0225-0,0075=0,015mol=n_{BaCO_3}\)(BT Ba)

\(\Rightarrow n_{CO_2}=n_{BaCO_3}=0,015mol\left(BtC\right)\)

Ta có:

\(\left\{{}\begin{matrix}100x+84y=1,42\\x+y=n_{CO_2}=0,015\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,01\\y=0,005\end{matrix}\right.\)

\(m_{CaCO_3}=0,01\cdot100=1g\)

\(m_{MgCO_3}=0,005\cdot84=0,42g\)

\(n_{CaCO_3}=\dfrac{40}{100}=0,4\left(mol\right)\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ n_{CO_2}=n_{CaCO_3}=0,4\left(mol\right)\\ \Rightarrow n_{NaOH}=0,4.2=0,8\left(mol\right)\\ m_{NaOH}=0,8.40=32\left(g\right)\\ \Rightarrow m_{ddNaOH}=\dfrac{32.100}{25}=128\left(g\right)\\ \Rightarrow V_{ddNaOH}=\dfrac{m_{ddNaOH}}{D_{ddNaOH}}=\dfrac{128}{1,28}=100\left(ml\right)=0,1\left(l\right)\)

a) Gọi x,y lần lượt là số mol Fe, Cu trong hhX (x,y>0) (mol)

- Khi cho X t/d hoàn toàn với khí Clo dư:

\(2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ Cu+Cl_2\rightarrow\left(t^o\right)CuCl_2\\ \Rightarrow162,5x+135y=59,5\left(1\right)\)

- Khi cho X tác dụng hoàn toàn với dd HCl 36,5%. Cu sẽ không tác dụng mà chỉ có Fe tham gia phản ứng.

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ m_{FeCl_2}=127x=25,4\left(g\right)\left(2\right)\\ \left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}162,5x+135y=59,5\\127x=25,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\\ \Rightarrow a=m_{hhX}=m_{Fe}+m_{Cu}=64x+56y=64.0,2+56.0,2=24\left(g\right)\)

Tính phần trăm mỗi muối sau phản ứng chắc ở phản ứng với Clo dư.

\(\%m_{FeCl_3}=\dfrac{0,2.162,5}{0,2.162,5+0,2.135}.100\approx54,622\%\\ \Rightarrow\%m_{CuCl_2}\approx45,378\%\)

b)

\(n_{HCl}=2x=2.0,2=0,4\left(mol\right)\\ \Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\\ \Rightarrow m_{ddHCl}=\dfrac{14,6.100}{36,5}=40\left(g\right)\\ \Rightarrow V_{ddHCl}=\dfrac{m_{ddHCl}}{D_{ddHCl}}=\dfrac{40}{1,25}=32\left(ml\right)=0,032\left(l\right)\)