Trong tinh thể : 

n Na2CO3 = n Na2CO3.10H2O = 5,72/286 = 0,02(mol)

Trong 200 gam dd Na2CO3 10% có :

m Na2CO3 = 200.10% = 20(gam)

Khi hòa tan tinh thể vào dung dịch trên : 

m Na2CO3 = 0,02.106 + 20 = 22,12(gam)

m dd = 5,72 + 200 = 205,72(gam)

C% Na2CO3 = 22,12/205,72  .100% = 10,75%

Sửa đề cho dễ làm: ^{_{"200g dd HCl 3,65%"}}

Ta có: \(n_{Na_2CO_3.10H_2O}=\dfrac{14,3}{106+10\cdot18}=0,05\left(mol\right)\) \(\Rightarrow n_{Na_2CO_3}=0,05\left(mol\right)\)

PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Na_2CO_3}=0,05\left(mol\right)\\n_{HCl}=\dfrac{200\cdot3,65\%}{36,5}=0,2\left(mol\right)\end{matrix}\right.\) 

Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{0,2}{2}\) \(\Rightarrow\) Na₂CO₃ p/ứ hết, HCl còn dư

\(\Rightarrow n_{NaCl}=0,1\left(mol\right)=n_{HCl\left(dư\right)}\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,1\cdot58,5=5,85\left(g\right)\\m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\end{matrix}\right.\)

Mặt khác: \(n_{CO_2}=0,05\left(mol\right)\) \(\Rightarrow m_{CO_2}=0,05\cdot44=2,2\left(g\right)\)

\(\Rightarrow m_{dd}=m_{Na_2CO_3.10H_2O}+m_{ddHCl}-m_{CO_2}=212,1\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{5,85}{212,1}\cdot100\%\approx2,76\%\\C\%_{HCl\left(dư\right)}=\dfrac{3,65}{212,1}\cdot100\%\approx1,72\%\end{matrix}\right.\)

\(Đặt:n_{Na_2CO_3}=a\left(mol\right);n_{K_2CO_3}=b\left(mol\right)\\ Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,3\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}106a+138b=38,2\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\\ a.C\%_{ddHCl}=\dfrac{0,6.36,5}{200}.100=10,95\%\\ b.m_{ddB}=38,2+200-0,3.44=225\left(g\right)\\ C\%_{ddKCl}=\dfrac{74,5.2.0,2}{225}.100\approx13,244\%\\ C\%_{ddNaCl}=\dfrac{58,5.2.0,1}{225}.100=5,2\%\)

trả lời giùm mik vs

https://hoc24.vn/hoi-dap/question/481532.html

\(n_{Na_2CO_3}=\dfrac{21,6}{106}\)

Số lẻ lắm em, em xem 21,6 hay 21,2 gam nhé!

a) 

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

PTHH: Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

               0,15<---------0,3<------------------------0,15

=> \(C\%_{dd.CH_3COOH}=\dfrac{0,3.60}{200}.100\%=9\%\)

b)

\(m_{dd.Na_2CO_3}=\dfrac{0,15.106.100}{15}=106\left(g\right)\)

c)

PTHH: 2CH₃COOH + Ba(OH)₂ --> (CH₃COO)₂Ba + 2H₂O

                    0,3--------->0,15

=> \(V_{dd.Ba\left(OH\right)_2}=\dfrac{0,15}{0,5}=0,3\left(l\right)=300\left(ml\right)\)