\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,5}{0,1}=5\left(M\right)\)

\(n_{SO_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

\(n_{KOH}=0,4.1=0,4\left(mol\right)\)

Ta có: \(\dfrac{n_{KOH}}{n_{SO_2}}=\dfrac{0,4}{0,3}=1,33\)

→ Pư tạo 2 muối: KHSO₃ và K₂SO₃

PT: \(SO_2+KOH\rightarrow KHSO_3\)

\(SO_2+2KOH\rightarrow K_2SO_3+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{SO_2}=n_{KHSO_3}+n_{K_2SO_3}=0,3\\n_{KOH}=n_{KHSO_3}+2n_{K_2SO_3}=0,4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{KHSO_3}=0,2\left(mol\right)\\n_{K_2SO_3}=0,1\left(mol\right)\end{matrix}\right.\)

m _{dd sau pư} = m_SO2 + m _ddKOH = 0,3.64 + 400.0,8 = 339,2 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{KHSO_3}=\dfrac{0,2.120}{339,2}.100\%\approx7,1\%\\C\%_{K_2SO_3}=\dfrac{0,1.158}{339,2}.100\%\approx4,7\%\end{matrix}\right.\)

Nồng độ phần trăm C% của dung dịch Y là khoảng 8,93%.

Ta có: \(n_{CO_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=0,15.1=0,15\left(mol\right)\)

\(\Rightarrow\dfrac{n_{CO_2}}{n_{Ca\left(OH\right)_2}}=\dfrac{0,2}{0,15}=1,33\)

→ Pư tạo 2 muối: Ca(HCO₃)₂ và CaCO₃.

PT: \(2CO_2+Ca\left(OH\right)_2\rightarrow Ca\left(HCO_3\right)_2\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=2n_{Ca\left(HCO_3\right)_2}+n_{CaCO_3}=0,2\\n_{Ca\left(OH\right)_2}=n_{Ca\left(HCO_3\right)_2}+n_{CaCO_3}=0,15\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Ca\left(HCO_3\right)_2}=0,05\left(mol\right)\\n_{CaCO_3}=0,1\left(mol\right)\end{matrix}\right.\)

m _{dd sau pư} = m_CO2 + m _ddCa(OH)2 - m_CaCO3 = 0,2.44 + 150.1,02 - 0,1.100 = 151,8 (g)

\(\Rightarrow C\%_{Ca\left(HCO_3\right)_2}=\dfrac{0,05.162}{151,8}.100\%\approx5,3\%\)

pthh: \(Fe+2HCl--->FeCl_2+H_2\uparrow\)

a. \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(\Rightarrow n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)

b. \(n_{FeCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

c. \(n_{HCl}=2.0,2=0,4\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)

`n_(Fe)=m/M=11,2/56=0,2(`mol`)`

PTHH: Fe + 2HCl `->` FeCl₂ + H₂

mol    : 1       2              1          1

mol    : 0,2     0,4         0,2       0,2

`V_(H_2)=n.24,79=0,2.24,79=4,958(`L`)`

`m_(FeCl_2)=n.M=0,2.127=25,4(`m`)`

`m_(HCl)=n.M=0,4.36.5=14,6`(g`)`

a. PTHH: \(Cuo+H_2SO_4\rightarrow CúSO_4+H_2\)

                 0,255    0,255          0,255         0,255

khối lượng chất tan H₂SO₄ là:

\(C\%=\dfrac{m_{ct}}{m_{đd}}\cdot100\%\Rightarrow m_{ct}=\dfrac{C\cdot m_{dd}}{100}\)

\(m_{ct}=\dfrac{20\cdot125}{100}=25\left(g\right)\)

số mol H₂SO₄ là: \(n_{H_2SO_4}=\dfrac{m}{M_{H2SO4}}=\dfrac{25}{98}\approx0,255\left(mol\right)\)

khối lượng của CuO là: \(m_{CuO}=n_{CuO}\cdot M_{CuO}=0,255\cdot80=20,4\left(g\right)\)

b. khối lượng CuSO4 tạo thành là:

\(m_{CuSO_4}=n_{CuSO_4}\cdot M_{CuSO_4}=0,255\cdot160=40,8\left(g\right)\)

\(4K+O_2\rightarrow2K_2O\\ 4Na+O_2\rightarrow2Na_2O\\ 2Ca+O_2\rightarrow2CaO\\ 4Al+3O_2\rightarrow2Al_2O_3\\ 2Fe+O_2\rightarrow2FeO\\ 2Cu+O_2\rightarrow2CuO\\ 4Ag+O_2\rightarrow2Ag_2O\\ S+O_2\rightarrow SO_2\\ C+O_2\rightarrow CO_2\\ 4P+5O_2\rightarrow2P_2O_5\)