\(n_{Na}=0.02\left(mol\right)\)

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

\(0.02....................0.02........0.01\)

\(V_{H_2}=0.01\cdot22.4=0.224\left(l\right)\)

\(m_{NaOH}=0.02\cdot40=0.8\left(g\right)\)

\(C\%_{NaOH}=\dfrac{0.8}{0.46+200-0.01\cdot2}\cdot100\%=0.4\%\)

\(n_{Ca} = a(mol)\\ Ca + 2H_2O \to Ca(OH)_2 + H_2\\ n_{H_2} = n_{Ca(OH)_2} = n_{Ca} = a(mol)\\ m_{dd\ sau\ pư} = 40a + 200 - 2a = 200 + 38a(gam)\\ C\%_{Ca(OH)_2} = \dfrac{74a + 200.1\%}{200 + 38a}.100\% = 2\%\\ \Rightarrow a = \dfrac{50}{1831} \to m_{Ca} = \dfrac{2000}{1831} =1,09(gam)\)

Ta có: \(C\%=20\%=\dfrac{m_{NaCl}}{250}.100\%\)

\(\Rightarrow m_{NaCl}=50\left(g\right)\)

\(m_{NaCl}=250.20\%=50\left(g\right)\)

Đề câu 2 sao khỏi làm đi

\(FeS+2HCl\rightarrow FeCl_2+H_2S\)\(\uparrow\)

 0.2                                    0.2

\(H_2S+4H_2O\rightarrow H_2SO_4+4H_2\)

0.2                        0.2          0.8

a. \(n_{FeS}=\dfrac{17.6}{88}=0.2mol\)

\(mdd_{H_2SO_4}=m_X=m_{H_2S}+m_{H_2O}-m_{H_2}=0.2\times34+92.3-0.8\times2=97.5g\)

\(C\%_{H_2SO_4}=\dfrac{0.2\times98\times100}{97.5}=20,1\%\)

b. \(\dfrac{1}{2}dd_X\Rightarrow n_X=0.1mol\)

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

  0.2              0.1

\(mdd_{NaOH}=\dfrac{0.2\times40\times100}{20}=40g\)

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)

Ta có \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(m_{NaOH}=100.16\%=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)

PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(Na_2O+H_2O\rightarrow2NaOH\)

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

\(n_{NaOH}=n_{Na}+2n_{Na_2O}\Rightarrow n_{Na_2O}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,2.23}{0,2.23+0,1.62}.100\%\approx42,6\%\\\%m_{Na_2O}\approx57,4\%\end{matrix}\right.\)