chắc là 8 lít SO2 không bạn, và ở điều kiện gì

\(n_{NaOH}=0.5\cdot0.5=0.25\left(mol\right)\)

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

\(0.25.........0.125...........0.125\)

\(C_{M_{Na_2CO_3}}=\dfrac{0.125}{0.5}=0.25\left(M\right)\)

a) \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

b) \(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Theo PT: \(n_{NaOH}=2n_{CO_2}=0,1\left(mol\right)\)

=> \(CM_{NaOH}=\dfrac{0,1}{0,1}=1M\)

c) Sửa đề D_NaOH = 1,2g/ml

 \(m_{ddsaupu}=0,05.44+100.1,2=122,2\left(g\right)\)

\(n_{Na_2CO_3}=n_{CO_2}=0,05\left(mol\right)\)

=> \(C\%_{Na_2CO_3}=\dfrac{0,05.106}{122,2}.100=4,34\%\)

\(n_{NaOH}=\dfrac{16.10\%}{40}=0,04\left(mol\right)\\ 2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\\ NaOH+SO_2\rightarrow NaHSO_3\\ Đặt:n_{NaHSO_3}=a\left(mol\right);n_{Na_2SO_3}=1,5a\left(mol\right)\\ \Rightarrow n_{NaOH\left(tổng\right)}=3a+a=4a=0,04\left(mol\right)\\ \Leftrightarrow a=0,01\left(mol\right)\\ n_{SO_2\left(tổng\right)}=n_{Na_2SO_3}+n_{NaHSO_3}=2,5a=0,025\left(mol\right)\\ V_{SO_2\left(đktc\right)}=0,025.22,4=0,56\left(lít\right)\)

Gọi $n_{Na_2CO_3} = a(mol) \Rightarrow n_{NaHCO_3} = 1,4a(mol)$
$2NaOH + CO_2 \to Na_2CO_3 + H_2O$
$NaOH + CO_2 \to NaHCO_3$
$n_{NaOH} = 2a + 1,4a = 3,4.0,5 \Rightarrow a = 0,5$

Suy ra :

$n_{CO_2} = a + 1,4a = 2,4a = 2,4.0,5 = 1,2(mol)$
$V_{CO_2} = 1,2.22,4 = 26,88(lít)$

a, \(n_{P_2O_5}=\dfrac{71}{142}=0,5\left(mol\right)\)

PT: \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

\(H_3PO_4+3NaOH\rightarrow Na_3PO_4+3H_2O\)

Theo PT: \(n_{H_3PO_4}=2n_{P_2O_5}=1\left(mol\right)\Rightarrow m_{H_3PO_4}=1.98=98\left(g\right)\)

b, Theo PT: \(n_{NaOH}=3n_{H_3PO_4}=3\left(mol\right)\Rightarrow m_{NaOH}=3.40=120\left(g\right)\)

c, \(n_{Na_3PO_4}=n_{H_3PO_4}=1\left(mol\right)\Rightarrow m_{Na_3PO_4}=1.164=164\left(g\right)\)

n NaOH=1,5(mol)

a) NaOH+ SO2----> NaHSO3

Để tạo muối axit thì \(\frac{n_{NaOH}}{n_{SO2}}\le1\)

\(\frac{1,5}{n_{SO2}}\le1\) => n\(_{SO2}\le1,5\)=> V\(_{SO2}\le33,6\left(l\right)\)

b)2 NaOH+SO2---->Na2SO3+H2O

Để tạo muối trung hòa thì \(\frac{n_{NaOH}}{n_{SO2}}\ge2\)

=> \(\frac{1,5}{n_{SO2}}\ge2\)=> \(n_{SO2}\ge0,75\Rightarrow V_{SO2}\ge16,8\left(l\right)\)

c) NaOH+SO2--->Na2SO3+H2O(1)

SO2+H2O+ Na2SO3----> 2NaHSO3(2)

Để tạo muối trung hòa thì

\(1< \frac{n_{KOH}}{n_{SO2}}< 2\) => \(1< \frac{1,5}{n_{SO2}}< 2\) =>

=> \(0,75< n_{SO2}< 1,5\)

=> \(16,8< V_{SO2}< 33,6\)

Vậy........

\(\text{n NaOH = 0.75x2 =1.5 mol}\)

Tạo muối axit => n SO3 < 1.5

Tạo muối trung hòa => n SO3 < 0.75

Tạo muối axit => n SO3 > 1.5

Tạo cả hai => 0.75< n SO3 < 1.5

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