Ta có: \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(n_{HCl}=0,4.1,5=0,6\left(mol\right)\)

Giả sử: \(\left\{{}\begin{matrix}n_{Na_2CO_3}=x\left(mol\right)\\n_{K_2CO_3}=y\left(mol\right)\end{matrix}\right.\)

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

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

\(HCl_{dư}+NaOH\rightarrow NaCl+H_2O\)

Theo PT: \(n_{CO_2}=n_{Na_2CO_3}+n_{K_2CO_3}=x+y\left(mol\right)\) ⇒ x + y = 0,25 (1)

\(n_{HCl\left(pư\right)}=2x+2y\left(mol\right)\) \(\Rightarrow n_{HCl\left(dư\right)}=0,6-2x-2y\left(mol\right)\)

Có: \(\left\{{}\begin{matrix}n_{NaCl}=2n_{Na_2CO_3}+n_{HCl\left(dư\right)}=0,6-2y\left(mol\right)\\n_{KCl}=2n_{K_2CO_3}=2y\left(mol\right)\end{matrix}\right.\)

⇒ 58,5(0,6 - 2y) + 74,5.2y = 39,9 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{0,1.106}{0,1.106+0,15.138}.100\%\approx33,9\%\\\%m_{K_2CO_3}\approx66,1\%\end{matrix}\right.\)

Bạn tham khảo nhé!

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%

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

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

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\Rightarrow m_{CO_2}=0,3.22,4=13,2\left(g\right)\)

PTHH: Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O

PTHH: K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O

PTHH: BaCO₃ + 2HCl → BaCl₂ + CO₂ + H₂O

PTHH: ZnCO₃ + 2HCl → ZnCl₂ + CO₂ + H₂O

Ta có: \(n_{HCl}=8n_{CO_2}=8.0,3=2,4\left(mol\right)\Rightarrow m_{HCl}=2,4.36,5=87,6\left(g\right)\)

         \(n_{H_2O}=n_{CO_2}=0,3\left(mol\right)\Rightarrow m_{H_2O}=5,4\left(g\right)\)

Theo ĐLBTKL ta có:

 \(m_{hh}+m_{HCl}=m_{muôi}+m_{CO_2}+m_{H_2O}\)

\(\Leftrightarrow m_{muối}=43,45+87,6-13,2-5,4=112,45\left(g\right)\)

nHCl (ban đầu) = 0.6(mol)
gọi nK2CO3 = x (mol)
nNa2CO3 = y(mol)
Ta có : x + y = nCO2 = 0.25(mol)
K2CO3 + 2HCl ---> 2KCl + H2O + CO2
x________2x____________________x
Na2CO3 + 2HCl ---> 2NaCl + H2O + CO2
y_________2y____________________y
=> n(HCl pư) = 2 x 0.25 = 0.5(mol)
=> nHCl dư = 0.1(mol) = nNaOH
=> dd thu được gồm : K(+) , Na(+), Cl(-) , Na(+)
Ta có : 78x + 46y = 39.9 - 0.6*35.5 - 23*0.1 = 16.3
=> x = 0.15
y = 0.1

BẠN THAM KHẢO

$n_{NaOH} = 0,2.2,5 = 0,5(mol)$
$BaCl_2 + Na_2CO_3 \to BaCO_3 + 2NaCl$
$n_{Na_2CO_3} = n_{BaCO_3} = \dfrac{39,4}{197} = 0,2(mol)$

TH1 : có tạo muối axit

$CO_2 + 2NaOH \to Na_2CO_3 + H_2O$
$CO_2 + NaOH \to NaHCO_3$

Suy ra:

$n_{NaHCO_3} = n_{NaOH}- 2n_{Na_2CO_3} = 0,5 - 0,2.2 = 0,1(mol)$
$n_{CO_2} = n_{Na_2CO_3} + n_{NaHCO_3} = 0,3(mol)$

$MgCO_3 + 2HCl \to MgCl_2 + CO_2 + H_2O$
$RCO_3 + 2HCl \to RCl_2 + CO_2 + H_2O$

$n_{MgCO_3} = n_{RCO_3} = \dfrac{0,3}{2} = 0,15(mol)$
Suy ra:

$0,15.84 + 0,15.(R + 60) = 20 \Rightarrow R = -10,6 \to$ Loại

TH2 : NaOH dư

$2NaOH + CO_2 \to Na_2CO_3 + H_2O$

$n_{CO_2} = n_{Na_2CO_3} = 0,2(mol)$
$n_{MgCO_3} = n_{RCO_3} = \dfrac{0,2}{2} = 0,1(mol)$
$\Rightarrow 0,1.84 + 0,1(R + 60) = 20 \Rightarrow R = 56(Fe)$

$\%m_{MgCO_3} = \dfrac{0,1.84}{20}.100\% = 42\%$

$\%m_{RCO_3} = 100\% -42\% = 58\%$

\(n_{CO2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

a) Pt : \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O|\)

            1              2             1            1          1

          0,1            0,2          0,1         0,1

           \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

             1           2              1               1

           0,05                      0,05

b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

⇒ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(m_{dd}=\dfrac{7,3.100}{7,3}=100\left(g\right)\)

d) \(n_{MgCO3}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{MgCO3}=0,1.84=8,4\left(g\right)\)

\(m_{MgO}=10,4-8,4=2\left(g\right)\)

\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)

\(n_{MgCl2}=0,1+0,05=0,15\left(mol\right)\)

⇒ \(m_{MgCl2}=0,15.95=14,25\left(g\right)\)

Sau phản ứng :

\(m_{dd}=10,4+100-\left(0,1.44\right)\)

       = 106 (g)

\(C_{MgCl2}=\dfrac{14,25.100}{106}=13,44\)⁰/₀

 Chúc bạn học tốt

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)

⇒ m_FeO = 12,6 - 5,4 = 7,2 (g)

c, Phần này đề cho dd NaOH dư hay vừa đủ bạn nhỉ?

d, Cho hh vào dd H₂SO₄ đặc nguội thì có khí thoát ra.

PT: \(2FeO+4H_2SO_{4\left(đ\right)}\rightarrow Fe_2\left(SO_4\right)_3+SO_2+4H_2O\)

Ta có: \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

Theo PT: \(n_{SO_2}=\dfrac{1}{2}n_{FeO}=0,05\left(mol\right)\)

\(\Rightarrow V_{SO_2}=0,05.22,4=1,12\left(l\right)\)