gọi số mol của NaOH = x , Ba(OH)2 = y . số mol OH- = x + 2y

pt : H+ + OH- → H2O 

nH+ =2. 0,035. 2=0,14 mol = x+ 2y 

khối lượng kết tủa = m_SO42-  + m_Ba2+ = 96.0.07 + 137.y= 9,32 =>y=0,02 => x= 0,1

nồng độ tự tính.

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1mol\\ CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\\ n_{BaCO_3}=n_{CO_2}=0,1mol\\ Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\\ n_{H_2SO_4}=n_{Ba\left(OH\right)_2}=0,1mol\\ m_{ddH_2SO_4}=\dfrac{0,1.98}{20\%}\cdot100\%=49g\\ V_{ddH_2SO_4}=\dfrac{49}{1,14}=42,98ml\)

\(n_{Ca\left(OH\right)_2}=0,2.0,4=0,08\left(mol\right)\)

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

              0,08---->0,16

=> \(C_{M\left(A\right)}=\dfrac{0,16}{0,3}=\dfrac{8}{15}M\)

\(n_{Ca\left(OH\right)_2}0,4.0,2=0,08\left(mol\right)\)

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

            0,016<-------------0,08

\(\rightarrow C_{M\left(A\right)}=\dfrac{0,04}{0,3}=0,533M\)

400ml = 0,4l

\(n_{Ca\left(OH\right)2}=0,2.0,4=0,08\left(mol\right)\)

Pt : \(2CH_3COOH+Ca\left(OH\right)_2\rightarrow\left(CH_3COO\right)_2Ca+2H_2O|\)

                 2                  1                         1                      2

               0,16              0,08

\(n_{CH3COOH}=\dfrac{0,08.2}{1}=0,16\left(mol\right)\)

300ml = 0,3l

\(C_{MddCH3COOH}=\dfrac{0,16}{0,3}=0,53\left(M\right)\)

 Chúc bạn học rốt

PTHH: \(2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\uparrow\)  (1)

            \(Na_2SO_4+Ba\left(OH\right)_2\rightarrow2NaOH+BaSO_4\downarrow\)  (2)

            \(2NaOH+MgCl_2\rightarrow2NaCl+Mg\left(OH\right)_2\downarrow\)  (3)

            \(Mg\left(OH\right)_2\xrightarrow[]{t^o}MgO+H_2O\)  (4)

Ta có: \(\left\{{}\begin{matrix}n_{Na_2SO_4}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2}\cdot\dfrac{2,3}{23}=0,05\left(mol\right)\\n_{BaCl_2}=\dfrac{60\cdot14,25\%}{208}=0,05\left(mol\right)\\n_{MgCl_2}=\dfrac{30\cdot19\%}{95}=0,06\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\) PT (2) p/ứ hết; PT (3) có MgCl₂ dư 0,01 mol  

\(\Rightarrow n_{MgO}=n_{Mg\left(OH\right)_2}=n_{BaSO_4}=0,05\left(mol\right)\)

\(\Rightarrow m_{rắn}=m_{MgO}+m_{BaSO_4}=0,05\cdot\left(40+233\right)=13,65\left(g\right)\)

b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=n_{Na}=0,1\left(mol\right)\\n_{Mg\left(OH\right)_2}=0,05\left(mol\right)=n_{H_2SO_4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,1\cdot58,5=5,85\left(g\right)\\m_{ddH_2SO_4}=\dfrac{0,05\cdot98}{4,9\%}=100\left(g\right)\\m_{Mg\left(OH\right)_2}=0,05\cdot58=2,9\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{Na}+m_{ddH_2SO_4}+m_{ddBaCl_2}+m_{ddMgCl_2}-m_{BaSO_4}-m_{Mg\left(OH\right)_2}=177,75\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{5,85}{177,75}\cdot100\%\approx3,29\%\\C\%_{MgCl_2\left(dư\right)}=\dfrac{0,01\cdot95}{177,75}\cdot100\%\approx0,53\%\end{matrix}\right.\) 

$Ba(OH)_2 + 2HNO_3 \to Ba(NO_3)_2 + 2H_2O$
$V_{dd\ sau\ pư} = 0,2 + 0,3 = 0,5(lít)$
$n_{Ba(NO_3)_2} = n_{Ba(OH)_2} = 0,2.0,2 = 0,04(mol)$
$C_{M_{Ba(NO_3)_2}} = \dfrac{0,04}{0,5} = 0,08M$

a, \(\left\{{}\begin{matrix}n_{Ba^{2+}}=4.10^{-3}\left(mol\right)\\n_{Na^+}=3.10^{-3}\left(mol\right)\\n_{OH^-}=0,011\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=\dfrac{4.10^{-3}}{0,2+0,3}=0,008M\\\left[Na^+\right]=\dfrac{3.10^{-3}}{0,2+0,3}=0,006M\\\left[OH^-\right]=\dfrac{0,011}{0,2+0,3}=0,022M\end{matrix}\right.\)

b, Để trung hòa dung dịch A thì:

\(n_{H^+}=n_{OH^-}\)

\(\Leftrightarrow0,01.V_{ddHCl}=\left(0,02.2+0,01\right).0,2\)

\(\Leftrightarrow V_{ddHCl}=1\left(l\right)\)

cần lời giải chi tiết ạ