MnO 2  + HCl →  MnCl 2  +  Cl 2  + 2 H 2 O

Cl 2  + 2NaOH → NaCl + NaClO +  H 2 O

n MnO 2 = 0,2 mol;   n NaOH = 0,729 mol

Theo phương trình (1) ta có: n Cl 2 = n MnO 2  = 0,2 mol

Theo phương trình (2) ta có:  2 n Cl 2 < n NaOH  ⇒ NaOH dư

Dung dịch A gồm: n NaCl = n NaClO = n Cl 2  = 0,2 mol

n NaOH   dư  = 0,729 – 2.0,2 = 0,329 mol

m dd   A = m Cl 2 + m dd   NaOH  = 0,2.71 + 145,8 = 160g

Chọn đáp án C

M n O 2 + 4HCl → M n C l 2 + C l 2 + 2 H 2 O

3 C l 2 + 6NaOH  → t 0 5NaCl + N a C l O 3 + 3 H 2 O

Phản ứng xảy ra:

\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)

Ta có:

\(n_{MnO2}=n_{Cl2}=\frac{34,8}{55+16.2}=0,4\left(mol\right)\)

Dẫn hỗn hợp khí vào NaOH

\(m_{NaOH}=291,6.20\%=58,32\left(g\right)\Rightarrow n_{NaOH}=\frac{58,32}{40}=1,458\left(mol\right)\)

\(\Rightarrow n_{NaOH}>2n_{Cl2}\) nên NaOH dư.

\(n_{NaCl}=n_{NaClO}=0,4\left(mol\right)\)

\(n_{NaOH\left(dư\right)}=1,458-0,4.2=0,658\left(mol\right)\)

BTKL:

\(m_{dd\left(A\right)}=m_{dd\left(NaOH\right)}+m_{Cl2}=291,6+0,4.71=320\left(g\right)\)

\(m_{NaCl}=0,4.58,5=23,4\left(g\right)\)

\(m_{NaClO}=0,4.74,5=29,8\left(g\right)\)

\(m_{NaOH\left(dư\right)}=0,658.40=26,32\left(g\right)\)

\(\Rightarrow C\%_{NaCl}=7,3125\%;C\%_{NaClO}=9,3125\%;C\%_{NaOH\left(dư\right)}=\frac{26,32}{320}=8,225\%\)

nhớ tick mình

chúc bn học tốt

\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)

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

Ta có :

\(n_{MnO2}=\frac{17,4}{87}=0,2\left(mol\right)\Rightarrow n_{Cl2}=0,2\left(mol\right)\)

\(n_{NaOH}=\frac{145,8.20\%}{40}=0,729\left(mol\right)\)

\(n_{NaOH.trong,A}=0,729-0,2.2=0,329\left(mol\right)\)

\(m_{dd\left(spu\right)}=0,2.71+145,8=160\left(g\right)\)

\(C\%_{NaOH}=\frac{0,329.40}{160}.100\%=8,225\%\)

\(C\%_{NaCl}=\frac{0,5.58,5}{160}.100\%=7,3125\%\)

\(C\%_{NaClO}=\frac{0,2.74,5}{160}.100\%=9,3125\%\)

\(n_{CaCO_3}=\dfrac{45}{100}=0,45\left(mol\right)\\ n_{NaOH}=0,5.1,5=0,75\left(mol\right)\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ n_{CO_2}=n_{CaCO_3}=0,45\left(mol\right)\\ Vì:1< \dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,75}{0,45}=\dfrac{5}{3}< 2\\ \Rightarrow Sp:Na_2CO_3,NaHCO_3\\ Đặt:n_{Na_2CO_3}=a\left(mol\right);n_{NaHCO_3}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}2a+b=0,75\\a+b=0,45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,15\end{matrix}\right.\\ \Rightarrow ddX:m_{Na_2CO_3}=0,3.106=31,8\left(g\right);m_{NaHCO_3}=84.0,15=12,6\left(g\right)\)

Số mol MnO₂ = 3,48 : 87 = 0,04 mol

Số mol NaOH = 0,1.0,8 = 0,08 mol

Ptpư: MnO₂ + 4HCl ---> MnCl₂ + Cl₂­ + 2H₂O

Cl₂ + 2NaOH---> NaCl + NaClO + H₂O

Theo ptpư: mol NaOH = 2mol Cl₂ ---> phản ứng vừa đủ

mol NaCl = mol NaClO = mol Cl₂ = 0,04 mol

--->C_M(NaCl) = C_M(NaClO) = 0,04 : 0,8 = 0,05(M)

Đáp án A

Chọn đáp án A

n M n O 2 = 69 , 6 87 =0,8 (mol); n N a O H = 0,5.4 = 2 (mol)

M n O 2 + 4HCl → M n C l 2 + C l 2 + 2 H 2 O

0,8                                    0,8      (mol)

C l 2 + 2NaOH → NaCl + NaClO + H 2 O

 0,8       1,6             0,8                 (mol)

C M   N a C l = 0 , 8 0 , 5  = 1,6(M); C M   ( N a O H   d ư ) = 2 - 1 ; 6 0 ; 5  = 0,8(M)