\(n_{Cl_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\\ C_{MddNaOH\left(dư\right)}=0,05\left(mol\right)\Rightarrow Tính.theo.Cl_2\\ n_{NaOH\left(P.Ứ\right)}=2.n_{Cl_2}=2.0,05=0,1\left(mol\right)\\ n_{NaOH\left(dư\right)}=0,2.0,05=0,01\left(mol\right)\\ \Rightarrow C_{MddNaOH\left(ban.đầu\right)}=\dfrac{0,1+0,01}{0,2}=0,55\left(M\right)\\ \Rightarrow Chọn.D\)

Câu 10:

\(PCl_3+3H_2O\rightarrow H_3PO_3+3HCl\\ H_3PO_3+3NaOH\rightarrow Na_3PO_3+3H_2O\\HCl+NaOH\rightarrow NaCl+H_2O\\ Đặt:n_{PCl_3}=k\left(mol\right)\\ \Rightarrow n_{H_3PO_3}=k\left(mol\right);n_{HCl}=3k\left(mol\right)\\ Ta.có:n_{NaOH}=3.n_{H_3PO_3}+n_{HCl}=3k+3k\\ \Leftrightarrow0,6=6k\\ \Leftrightarrow k=0,1\left(mol\right)\\ \Rightarrow n_{PCl_3}=0,1\left(mol\right)\\ \Rightarrow Chọn.A\)

Câu 8:

\(n_{NaCl}=\dfrac{8,775}{58,5}=0,15\left(mol\right)\\ NaCl+H_2SO_{4\left(đặc\right)}\rightarrow\left(250^oC\right)NaHSO_4+HCl\uparrow\\ n_{HCl}=n_{NaCl}=0,15\left(mol\right)\\ \Rightarrow C\%_{ddHCl}=\dfrac{0,15.36,5}{0,15.36,5+14,525}.100=27,375\%\\ \Rightarrow Chọn.A\\ MnO_2+4HCl_{đặc}\rightarrow\left(t^o\right)MnCl_2+Cl_2+2H_2O\\ n_{Cl_2\left(thu\right)}=70\%.n_{Cl_2\left(LT\right)}=70\%.\dfrac{0,15}{4}=0,02625\left(mol\right)\\ \Rightarrow V_{Cl_2\left(đktc,thu\right)}=0,02625.22,4=0,588\left(l\right)\\ \Rightarrow Chọn.B\)

\(2M+2H_2O\rightarrow2MOH+H_2\\ M_2O+H_2O\rightarrow2MOH\\ Đặt:n_M=a\left(mol\right);n_{M_2O}=b\left(mol\right)\\ MOH+HCl\rightarrow MCl+H_2O\\ n_{HCl}=0,2.0,35=0,07\left(mol\right)\\ \Rightarrow n_{MOH}=0,07\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}a.M_M+b.\left(2M_M+16\right)=1,93\\0,5a=0,015\\a+2b=0,07\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,03\\b=0,02\\M_M=23\left(\dfrac{g}{mol}\right)\end{matrix}\right.\\ \Rightarrow M\left(I\right):Natri\left(Na=23\right)\)

=>Chọn B

thank u

1)

Theo pt:

-1mol CaOCl2 sẽ tạo ra 1mol Cl2

-1mol KMnO4 sẽ tạo ra 2,5mol Cl2

-1mol K2Cr2O7 sẽ tạo ra 3mol Cl2

-1mol MnO2 sẽ tạo ra 1mol Cl2

Vậy K2Cr2O7 tạo ra lượng khí clo nhiều nhất,CaOCl2 và MnO2 tạo ra lượng khí clo ít nhất

Theo đề, ta có: \(n_{CaOCl_2}=\dfrac{100}{127}\left(mol\right)\\ n_{KMnO_4}=\dfrac{100}{158}=\dfrac{50}{79}\left(mol\right)\\ n_{K_2Cr_2O_7}=\dfrac{100}{294}=\dfrac{50}{147}\left(mol\right)\\ n_{MnO_2}=\dfrac{100}{87}\left(mol\right)\)|

\(CaOCl_2+2HCl\rightarrow Cl_2+H_2O+CaCl_2\)

\(16HCl+2KMnO_4\rightarrow5Cl_2+8H_2O+2KCl+2MnCl_2\)

\(14HCl+K_2Cr_2O_7\rightarrow3Cl_2+7H_2O+2KCl+2CrCl_3\)

\(4HCl+MnO_2\rightarrow MnCl_2+2H_2O\)

Theo phương trình:

- \(\dfrac{100}{127}mol\) CaOCl₂ sẽ tạo ra \(\dfrac{100}{127}mol\) Cl₂.

- \(\dfrac{100}{158}mol\) KMnO₄ sẽ tạo ra \(\dfrac{125}{79}mol\) Cl₂.

- \(\dfrac{100}{294}mol\) K₂Cr₂O₇ sẽ tạo ra \(\dfrac{50}{49}mol\) Cl₂.

- \(\dfrac{100}{87}mol\) MnO₂ sẽ tạo ra \(\dfrac{100}{87}mol\) Cl₂.

Vậy 100(g) KMnO₄ sẽ tạo ra lượng Cl₂ nhiều nhất, 100(g) CaOCl₂ sẽ tạo ra lượng Cl₂ ít nhất.

\(n_{halogen}=\dfrac{35,875-18,625}{108-39}=0,25\left(mol\right)\\ M_{muối\left(t.gia\right)}=\dfrac{18,625}{0,25}=74,5=39+M_{halogen}\\ \Leftrightarrow M_{halogen}=35,5\left(\dfrac{g}{mol}\right)\\ Vậy.muối.halogen.đó:KCl\left(Cl=35,5\right)\\ Chọn.A\)

Câu 29:

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

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

(3) \(H_2+Cl_2\underrightarrow{t^o}2HCl\)

(4) \(HCl+NaOH\rightarrow NaCl+H_2O\)

(5) \(NaCl+AgNO_3\rightarrow NaNO_3+AgCl_{\downarrow}\)

(6) \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Câu 30:

a, PT: \(Zn+S\underrightarrow{t^o}ZnS\)

\(Fe+S\underrightarrow{t^o}FeS\)

b, Giả sử: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

⇒ 65x + 56y = 26,55 (1)

Ta có: \(n_S=\dfrac{14,4}{32}=0,45\left(mol\right)\)

Theo PT: \(n_S=n_{Zn}+n_{Fe}=x+y\left(mol\right)\)

⇒ x + y = 0,45 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,15\left(mol\right)\\y=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,15.65}{26,55}.100\%\approx36,7\%\\\%m_{Fe}\approx63,3\%\end{matrix}\right.\)

Bạn tham khảo nhé!