\(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right);n_{NaOH}=0,2.1,75=0,35\left(mol\right)\)

Ta có: \(T=\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,35}{0,15}=\dfrac{7}{3}\approx2,33\) ⇒ pt tạo ra muối Na₂CO₃

PTHH: CO₂ + 2NaOH → Na₂CO₃ + H₂O

Mol:     0,15       0,3             0,15    

\(C_{M_{ddNa_2CO_3}}=\dfrac{0,15}{0,2}=0,75M\)

\(C_{M_{ddNaOHdư}}=\dfrac{0,35-0,3}{0,2}=0,25M\)

PTHH: CO₂ + NaOH ---> NaHCO₃

Đổi 200ml = 0,2 lít

Ta có: \(V_{dd_{NaHCO_3}}=3,36+0,2=3,56\left(lít\right)\)

Ta có: \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{NaHCO_3}=n_{CO_2}=0,15\left(mol\right)\)

=> \(C_{M_{NaHCO_3}}=\dfrac{0,15}{3,56}=0,04M\)

\(n_{Cl_2}=\dfrac{3,36}{22,4}=0,15mol\\ Cl_2+2NaOH\rightarrow NaCl+NaOH+H_2O\\ n_{NaCl}=n_{NaOH}=n_{Cl_2}=0,15mol\\ C_{M_{NaCl}}=\dfrac{0,15}{0,3}=0,5M\\ C_{M_{NaClO}}=\dfrac{0,15}{0,3}=0,5M\)

\(n_{Cl_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

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

 2 mol        1 mol    1mol      1mol      1 mol

0,1          0,05         0,05         0,05      0,05

\(V_{NaOH}=\dfrac{0,1}{1}=0,1\left(l\right)\)

\(CM_{NaCl}=\dfrac{0,05}{0,1}=0,5M\)

\(CM_{NaClO}=\dfrac{0,05}{0,1}=0,5M\)

Chọn đáp án C

Chọn đáp án C

n C l 2 = 1 , 12 22 , 4  = 0,05 (mol); n N a O H d ư = 0,2.0,5 = 0,1 (mol)

C l 2 + 2NaOH → NaCl + NaClO + H 2 O

 0,05    0,1                     (mol)

=> n N a O H b đ = 0,1 + 0,1 = 0,2 (mol) => C M N a O H _­ =  = 1,0 (M)

Cl2 + 2 NaOH -> NaCl + NaClO + H2O

nNaOH= 0,7(mol); nCl2=0,3(mol)

Ta có: 0,7/2 > 0,3/1

=>Cl2 hết, NaOH dư, tính theo nCl2

-> dd sau p.ứ có NaCl , NaClO và NaOH(dư)

nNaCl=nNaClO=nCl2= 0,3(mol)

nNaOH(Dư)=0,7-0,3.2=0,1(mol)

=Vddsau=  VddNaOH=0,2(l)

=>CMddNaCl= 0,3/0,2=1,5(M)

CMddNaClO=0,3/0,2=1,5(M)

CMddNaOH(dư)=0,1/0,2=0,5(M)

\(\left\{{}\begin{matrix}n_{H_2S}=\dfrac{2,464}{22,4}=0,11\left(mol\right)\\n_{NaOH}=0,05.2,5=0,125\left(mol\right)\end{matrix}\right.\)

Xét \(\dfrac{n_{NaOH}}{n_{H_2S}}=\dfrac{0,125}{0,11}=1,1\) => Tạo ra muối Na₂S, NaHS

Gọi số mol NaOH phản ứng là để tạo muối Na₂S, NaHS lần lượt là a, b (mol)

=> a + b = 0,125 (1)

PTHH: 2NaOH + H₂S --> Na₂S + 2H₂O 

                  a---->0,5a---->0,5a

            NaOH + H₂S --> NaHS + H₂O

                  b---->b---------->b

=> 0,5a + b = 0,11 (2)

(1)(2) => a = 0,03 (mol); b = 0,095 (mol)

=> \(\left\{{}\begin{matrix}C_{M\left(Na_2S\right)}=\dfrac{0,5.0,03}{0,05}=0,3M\\C_{M\left(NaHS\right)}=\dfrac{0,095}{0,05}=1,9M\end{matrix}\right.\)

Bài 7:

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\n_{NaOH}=0,2\cdot1=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo 2 muối

PTHH: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

                 a_______2a__________a              (mol)

            \(CO_2+NaOH\rightarrow NaHCO_3\)

                 b_______b__________b       (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}a+b=0,15\\2a+b=0,2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,1\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{CO_2}+m_{ddNaOH}=0,15\cdot44+200\cdot1,25=256,6\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2CO_3}=\dfrac{0,05\cdot106}{256,6}\cdot100\%\approx2,1\%\\C\%_{NaHCO_3}=\dfrac{0,1\cdot72}{256,6}\cdot100\%\approx2,8\%\end{matrix}\right.\)

Bài 8:

PTHH: \(RCO_3+2HNO_3\rightarrow R\left(NO_3\right)_2+CO_2\uparrow+H_2O\)

Giả sử \(n_{RCO_3}=1\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{HNO_3}=2\left(mol\right)\\n_{R\left(NO_3\right)_2}=1\left(mol\right)=n_{CO_2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ddHNO_3}=\dfrac{2\cdot63}{20\%}=630\left(g\right)\\m_{R\left(NO_3\right)_2}=R+124\left(g\right)\\m_{CO_2}=44\left(g\right)\end{matrix}\right.\) \(\Rightarrow C\%_{R\left(NO_3\right)_2}=\dfrac{124+R}{R+60+630-44}=0,26582\)

\(\Leftrightarrow R=65\) (Kẽm) \(\Rightarrow\) CTHH của muối cacbonat là ZnCO₃