a) \(n_{Na}=\dfrac{11,5}{23}=0,5\left(mol\right)\)

\(n_{NaOH}=\dfrac{8\%.500}{40}=1\left(mol\right)\)

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

0,5---------------->0,5------->0,25

\(\Sigma n_{NaOH}=0,5+1=1,5\left(mol\right)\)

\(m_{ddsaupu}=11,5+500-0,25.2=511\left(g\right)\)

=> \(C\%_{NaOH}=\dfrac{1,5.40}{511}.100=11,74\%\)

b) Gọi thể tích dung dịch X cần tìm là V

 \(n_{H^+}=V.1+V.0,5.1=2V\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

Ta có : \(n_{H^+}=n_{OH^-}=1,5\left(mol\right)\)

=> 2V=1,5

=> V=0,75(lít)

\(n_{BaSO_4}=\dfrac{23.3}{233}=0.1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+H_2O\)

\(n_{BaO}=n_{Ba\left(OH\right)_2}=n_{BaSO_4}=0.1\left(mol\right)\)

\(m_{BaO}=0.1\cdot153=15.3\left(g\right)\)

\(m_{Na_2O}=24.6-15.3=9.3\left(g\right)\)

\(n_{Na_2O}=\dfrac{9.3}{62}=0.15\left(mol\right)\)

\(\%BaO=62.2\%\)

\(\%Na_2O=37.8\%\)

\(2.\)

\(m_{ddX}=24.6+73.7=98.3\left(g\right)\)

\(n_{H_2SO_4}=\dfrac{0.15}{2}+0.1=0.175\left(mol\right)\)

\(m_{dd_{H_2SO_4}}=\dfrac{0.175\cdot98\cdot100}{19.6}=87.5\left(g\right)\)

\(m_{ddY}=m_{ddX}+m_{ddH_2SO_4}-m_{\downarrow}=98.3+87.5-23.3=162.5\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{0.075\cdot142}{162.5}\cdot100\%=6.55\%\)

B1:

2NaOH+H₂SO₄\(\rightarrow\)Na₂SO₄+2H₂O

n_NaOH=\(\frac{4}{40}=0.1\)mol

=>n_H2SO4=\(\frac{1}{2}\)n_NaOH=0.05 mol

=>C_M=\(\frac{n_{H2SO42}}{V}\)=\(\frac{0.05}{200}\)=2,5.10^-4 (M)

B2:

Mg+\(\frac{1}{2}\)O₂\(\underrightarrow{t^0}\)MgO                (1)

MgO+2HCl\(\rightarrow\)MgCl₂+H₂O (2)

n_Mg(1)=\(\frac{0,36}{24}=0,015mol\)

=>n_MgO(1)=0,015=n_MgO(2)

n_HCl(2)=2n_MgO(2)=0,03mol

=>C_M(HCl)=\(\frac{n_{HCl}}{V}=\frac{0,03}{100}=3.10^{-4}M\)

nK=0,25(mol)

2K+2H2O→2KOH+H2↑

Theo PT nKOH=nK=0,25(mol)

=> mKOH=0,25.56=14(g)

Theo PT nH2=0,5nK=0,5.0,25=0,125(mol)

=> mH2=0,125.2=0,25(g)

mdd=9,75+400−0,25=409,5(g)

mNaOH=400.15%=60(g)

C%KOH=\(\dfrac{14}{409,5}.100\)=3,419%

C%NaOH=\(\dfrac{60}{409,5}.100\)=14,652%

\(n_P=\dfrac{3.1}{31}=0.1\left(mol\right)\)

\(4P+5O_2\underrightarrow{^{^{t^0}}}2P_2O_5\)

\(0.1.................0.05\)

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

\(0.05............................0.1\)

\(C_{M_{H_3PO_4}}=\dfrac{0.1}{0.1}=1\left(M\right)\)

\(m_{H_3PO_4}=0.1\cdot98=9.8\left(g\right)\)

\(m_{dd}=0.05\cdot142+100=107.1\left(g\right)\)

\(C\%_{H_3PO_4}=\dfrac{9.8}{107.1}\cdot100\%=9.15\%\)

Ta có: \(m_{KCl}=558,75\cdot10\%+7,45=63,325\left(g\right)\)

\(\Rightarrow C\%_{KCl}=\dfrac{63,325}{558,75+7,45}\cdot100\%\approx11,2\%\)

\(n_{NaOH}=\dfrac{12.75}{40}=\dfrac{51}{160}\left(mol\right)\)

\(C_{M_{NaOH}}=\dfrac{\dfrac{51}{160}}{0.3}=1.0625\left(M\right)\)