a/ A+ HCl
CO3 2- + 2H+ ---> H2O+ CO2
dd B trung hòa bởi NaOH--> trong B có Ba(HCO3)2
CO2 + Ba(OH)2 --> BaCO3 + H2O
0.2<---0.25-0.05-------->0.2
2Co2+ Ba(OH)2--> Ba(HCO3)2
0.1<--------0.05<---------0.05
Ba(HCO3)2+ 2NaOH---> BaCO3+ Na2CO3+ 2H2O
0.05<-------------0.1
--> m2= 0.2*197=39,4g
Na2CO3 va K2CO3 : x,y mol

x+y=0.3
138y=106x*2,604
-->x=0.1,y=0.2
--> m1=0.1*106+ 0,2*138=38,2
b/

C%Na2CO3= (0.1*106*100)/ (61,8+ 38,2)=10,6%
C%K2CO3=(0.2*138*100)/(61,8+ 38,2)=27,6%

m giỏi thế

Ta có : \(n_{Ba\left(OH\right)2}=0,25\left(mol\right);n_{NaOH}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{BaCO3}=0,2\left(mol\right)\\n_{Ba\left(HCO3\right)2}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_2=39,4g\\m_{CO2}=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Na2CO3}+n_{K2CO3}=0,3\\2,604.106n_{Na2CO3}=138n_{K2CO3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{na2CO3}=0,1\left(mol\right)\\n_{K2CO3}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_1=106n_{Na2CO3}+138n_{K2CO3}=38,2g\)

\(\Rightarrow\left\{{}\begin{matrix}m_1=38,2g\\m_2=39,4g\end{matrix}\right.\)

b, Ta có :

2 muối trong dung dịch là :

Na2CO3 và K2CO3

\(\text{=> C% Na2CO3 = (0,1.106.100)/(61,8 + 38,2) = 10,6% }\)

\(\text{=> C% K2CO3 = (0,2.138.100)/(61,8 + 38,2) = 27,6% }\)

A+ HCl
CO3 2- + 2H+ ---> H2O+ CO2
dd B trung hòa bởi NaOH--> trong B có Ba(HCO3)2
CO2 + Ba(OH)2 --> BaCO3 + H2O
0.2<---0.25-0.05-------->0.2
2Co2+ Ba(OH)2--> Ba(HCO3)2
0.1<--------0.05<---------0.05
Ba(HCO3)2+ 2NaOH---> BaCO3+ Na2CO3+ 2H2O
0.05<-------------0.1
--> m2= 0.2.197=39,4g
Na2CO3 va K2CO3 : x,y mol

x+y=0.3
138y=106x.2,604
-->x=0.1,y=0.2
--> m1=0.1.106+ 0,2.138=38,2
b/C%Na2CO3= (0.1.106.100)/ (61,8+ 38,2)=10,6%
C%K2CO3=(0.2.138.100)/(61,8+ 38,2)=27,6%
c/CO3 2- + 2H+ ---> H2O+ CO2
0.3----------->0.6
[H+]=[HCl]=0.6/2=0.3
d// C% Na2CO3= 27,6/100=[ (x+0.1).106] /(61,8+ 38,2)
-->x=0.16--> nNa2CO3 cần thêm=16,96g

#Tham khảo

A+ HCl
CO3 2- + 2H+ ---> H2O+ CO2
dd B trung hòa bởi NaOH--> trong B có Ba(HCO3)2
CO2 + Ba(OH)2 --> BaCO3 + H2O
0.2<---0.25-0.05-------->0.2
2Co2+ Ba(OH)2--> Ba(HCO3)2
0.1<--------0.05<---------0.05
Ba(HCO3)2+ 2NaOH---> BaCO3+ Na2CO3+ 2H2O
0.05<-------------0.1
--> m2= 0.2*197=39,4g
Na2CO3 va K2CO3 : x,y mol

x+y=0.3
138y=106x*2,604
-->x=0.1,y=0.2
--> m1=0.1*106+ 0,2*138=38,2
b/C%Na2CO3= (0.1*106*100)/ (61,8+ 38,2)=10,6%
C%K2CO3=(0.2*138*100)/(61,8+ 38,2)=27,6%
c/CO3 2- + 2H+ ---> H2O+ CO2
0.3----------->0.6
[H+]=[HCl]=0.6/2=0.3
d// C% Na2CO3= 27,6/100=[ (x+0.1)*106] /(61,8+ 38,2)
-->x=0.16--> nNa2CO3 cần thêm=16,96g

Phùng Hà Châu,Ten Hoàng,Trần Hữu Tuyển....... giúp e vs ạ~~~~~~~

a) Gọi $n_{CO_2} = a(mol) ; n_{SO_2} = b(mol)$
Ta có : 

$a + b = \dfrac{8,96}{22,4} = 0,4(mol)$

$\dfrac{44a + 64b}{a + b} = 27.2$

Suy ra : a = b = 0,2$
$V_{CO_2} = V_{SO_2} = 0,2.22,4 = 4,48(lít)$

b) Theo PTHH : $n_{K_2SO_3} = n_{SO_2} = 0,2(mol)$

$\Rightarrow m_{K_2SO_3} = 0,2.158 = 31,6(gam)$

Gọi $n_{K_2CO_3} = x(mol) ; n_{Na_2CO_3} = y(mol)$

$\Rightarrow 138x + 106y + 31,6 = 56(1)$

$n_{CO_2} = x + y = 0,2(2)$

Từ (1)(2) suy ra : x = y = 0,1

$m_{K_2CO_3} = 0,1.138 = 13,8(gam) ; m_{Na_2CO_3} = 0,1.106 = 10,6(gam)$

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

\(448ml=0,448l\)

\(20ml=0,02l\)

\(n_{CO_2}=\frac{0,448}{22,4}=0,02mol\)

\(n_{HCl}=2n_{CO_2}=0,04mol\)

\(CMHCl=\frac{0,04}{0,02}=2M\)

b. \(n_{Na_2CO_3}=n_{CO_2}=0,02mol\)

\(\%m_{Na_2CO_3}=\frac{0,02.106}{5}.100\%=42,4\%\)

\(\%m_{NaCl}=100-42,4=57,6\%\)

Gọi số mol của Na2CO3 là a, K2CO3 là b, do HCl dư.

Vậy CO3 2-biến thành CO2

CO3 2-+ 2 H+-----> CO2+ H2O

a + b a + b
Ta có : a + b = 2,016/ 22,4 = 0,09 mol
106a + 138b = 10,5
giải hệ : a = 0,06 mol Na2CO3

cảm ơn bạn nhoa

PTHH :

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

x                  2x           2x            x           x 

\(K_2CO_3+2HCl\rightarrow2KCl+H_2O+CO_2\uparrow\)

y                  2y           2y          y            y

Ta có : 

106x + 138y = 26 

2x + 2y = 0,4 

Giải hệ PT, ta có :

\(\rightarrow x=0,05\left(mol\right);y=0,15\left(mol\right)\)

Thu đc khí CO2 chứ bạn nhỉ?

\(a,V_{CO_2}=\left(0,05+0,15\right).22,4=4,48\left(l\right)\)

\(b,m_{muối}=0,05.58,5+0,15.74,5=14,1\left(g\right)\)

\(c,\%m_{Na_2CO_3}=\dfrac{0,05.106}{26}.100\%\approx20,38\%\)

\(\%m_{K_2CO_3}=100\%-20,38\%=79,62\%\)

\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\\ BaCO_3+2HCl\rightarrow BaCl_2+CO_2+H_2O\\ ZnCO_3+2HCl\rightarrow ZnCl_2+CO_2+H_2O\\ n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{muối.khan}=43,45+0,3.\left(71-60\right)=46,75\left(g\right)\)

Ta có: \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\) \(\Rightarrow m_{CO_2}=0,3\cdot44=13,2\left(g\right)\)

Bảo toàn nguyên tố: \(n_{CO_2}=n_{H_2O}=\dfrac{1}{2}HCl=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{H_2O}=0,3\cdot18=5,4\left(g\right)\\m_{HCl}=0,3\cdot2\cdot36,5=21,9\left(g\right)\end{matrix}\right.\)

Bảo toàn khối lượng: \(m_{muối}=m_{hh\left(ban.đầu\right)}+m_{HCl}-m_{H_2O}-m_{CO_2}=46,75\left(g\right)\)