\(NaOH+HCl--->NaCl+H2O\)

\(Na2CO3+2HCl--.2NaCl+H2O+CO2\uparrow\)

\(n_{CO2}=\frac{4,48}{22,4}=0,1\left(mol\right)\)

\(n_{Na2CO3}=n_{CO2}=0,2\left(mol\right)\)

\(n_{Na2CO3}=106.0,2=21,2\left(mol\right)\)

\(\%\%m_{Na2CO3}=\frac{21,2}{30}.100\%=70,67\%\)

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

tl............1................2.............2.............1.............1..(mol

br     0,1.................0,2......................................0,1(mol)

NaCl không phản ứng đc vsHCl

b)\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(C_{MHCl}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)(đổi 400ml=0,4(l))

c)\(Tacom_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)

\(\Rightarrow\%m_{Na_2CO_3}=\dfrac{10,6}{20}.100=53\%\)

\(\Rightarrow\%mNaCl=100\%-53\%=47\%\)

PTHH:

Na2CO3 + 2HCl -----> 2NaCl + H2O + CO2   (1)

K2CO3 + 2HCl -----> 2KCl + H2O + CO2      (2)

NaOH + HCl ----> NaCl + H2O      (3)

Gọi n Na2CO3 = a  ,  n K2CO3 = b (mol)

Theo pt(1)(2) tổng n CO2= a+b=\(\frac{5,6}{22,4}\)=0,25  (I)

n HCl = 1,5 . 0,4= 0,6 (mol)

Theo pt(1)(2) tổng n HCl pư=2 (a+b)=0,5 (mol)

==> n HCl dư= 0,1 mol

Theo pt(3) n NaCl= n HCl=0,1 mol   ==> m NaCl=5,85 (g)

Theo pt(1)(2) n NaCl=2a  ==> m NaCl= 117a

                       n KCl=2b ==> m KCl= 149b

===> 117a + 149b + 5,85 = 39,9

-----> 117a + 149b = 34,05 (II)

Từ (I)và (II) ==> a=0,1 và b=0,15

==>m hh = 0,1 . 106 + 0,15 . 138= 31,3(g)

m Na2CO3=10,6 (g)

%m Na2CO3 = \(\frac{10,6}{31,3}\) . 100%= 33,87%

%m K2CO3 = 10% - 33,87% = 66,13%

448ml = 0,448l

\(n_{CO2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)

a) Pt : \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O|\)

                  1             2              2           1           1

                0,02                                    0,02

b) \(n_{Na2CO3}=\dfrac{0,02.1}{1}=0,02\left(mol\right)\)

\(m_{Na2CO3}=0,02.106=2,12\left(g\right)\)

\(m_{NaCl}=5-2,12=2,88\left(g\right)\)

c) ⁰/₀NaCl = \(\dfrac{2,88.100}{5}=57,6\)⁰/₀

     ⁰/₀Na₂CO₃ = \(\dfrac{2,12.100}{5}=42,4\)⁰/₀

 Chúc bạn học tốt

\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)

 \(x\)                  2x          2x            x           x

\(NaHCO_3+HCl\rightarrow NaCl+H_2O+CO_2\)

 \(y\)                  y            y            y           y

Ta có hệ:

\(\left\{{}\begin{matrix}106x+84y=19\\x+y=\dfrac{4,48}{22,4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

\(m_{Na_2CO_3}=0,1\cdot106=10,6g\)

\(m_{NaHCO_3}=0,1\cdot84=8,4g\)

\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ NaHCO_3+HCl\rightarrow NaCl+CO_2+H_2O\\ Đặt:n_{Na_2CO_3}=a\left(mol\right);n_{NaHCO_3}=b\left(mol\right)\left(a,b>0\right)\\ m_{hh.muối.ban.đầu}=19\left(g\right)\\ \Leftrightarrow106a+84b=19\left(1\right)\\ Mặt.khác:V_{CO_2\left(tổng\right)}=22,4a+22,4b=4,48\left(2\right)\\ \left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}106a+84b=19\\22,4a+22,4b=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \Rightarrow m_{NaHCO_3}=84.0,1=8,4\left(g\right)\\ m_{Na_2CO_3}=106.0,1=10,6\left(g\right)\)

Na 2 CO 3  + 2HCl → 2NaCl +  H 2 O + CO 2

n khi = n CO 2  = 0,448/22,4 = 0,02 mol;  n HCl  = 0,02.2/1 = 0,04 mol

n Na 2 CO 3  = 0,02.1/1 = 0,02 (mol) →  m Na 2 CO 3  = 0,02 x 106 = 2,12g

% m Na 2 CO 3  = 2,12/5 . 100% = 42,4%

% m NaCl  = 100% - 42,4% = 57,6%

Chọn D