\(n_{K_2O}=\dfrac{2,35}{94}=0,025\left(mol\right)\\ K_2O+H_2O\rightarrow2KOH\\ n_{KOH}=0,025.2=0,05\left(mol\right)\\ V_{ddKOH}=V_{ddH_2O}=400\left(ml\right)=0,4\left(l\right)\\ C_{MddKOH}=\dfrac{0,05}{0,4}=0,125\left(M\right)\)

400ml = 0,4l

\(n_{K2O}=\dfrac{2,35}{94}=0,025\left(mol\right)\)

Pt : \(K_2O+H_2O\rightarrow2KOH|\)

          1         1            2

       0,025                 0,05

\(n_{KOH}=\dfrac{0,025.2}{1}=0,05\left(mol\right)\)

\(C_{M_{ddKOH}}=\dfrac{0,05}{0,4}=0,125\left(M\right)\)

 Chúc bạn học tốt

K2O+H2O->2KOH

0,025-----------0,05 mol

n K2O=2,35\94=0,025 mol

=>CmKOH=0,05\0,4=0,125M

a) \(n_{K_2O}=\dfrac{37,6}{94}=0,4\left(mol\right)\)

PTHH: K₂O + H₂O --> 2KOH

              0,4------------>0,8

=> m_KOH = 0,8.56 = 44,8 (g)

b) \(C_M=\dfrac{0,8}{0,5}=1,6M\)

c) 

PTHH: 4K + O₂ --t^o--> 2K₂O

                     0,2<----0,4

=> V_O2 = 0,2.22,4 = 4,48 (l)

nKK 2O = 37,6\94 = 0,4 mol

K2O+H2O --> 2KOH

 0,4                0,4                   0,8        ( mol )

mKOH= 0,8.56=44,8g

CMMKOH = 0,80,50,80,5 = 1,6M

c. 

4K+O2 --to> 2K2O

0,4       0,1                                 ( mol )

VO2 = 0,1.22,4=2,24lít

Câu 1 : 

\(n_{K_2O}=\dfrac{2.35}{94}=0.025\left(mol\right)\)

\(K_2O+H_2O\rightarrow2KOH\)

\(0.025...................0.05\)

\(C_{M_{KOH}}=\dfrac{0.05}{0.4}=0.125\left(M\right)\)

Câu 2 : 

\(n_{Ca\left(OH\right)_2}=\dfrac{1.11}{111}=0.01\left(mol\right)\)

\(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

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

\(C_{M_{CaCl_2}}=\dfrac{0.01}{0.5}=0.02\left(M\right)\)

Câu 3 : 

\(n_{Al_2\left(SO_4\right)_3}=0.6\cdot0.05=0.03\left(mol\right)\)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

\(0.03...............................0.03\)

\(m_{Al_2O_3}=0.03\cdot102=3.06\left(g\right)\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

           \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

Ta có: \(\left\{{}\begin{matrix}\Sigma n_{HCl}=0,4\cdot2=0,8\left(mol\right)\\n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,1mol\\n_{Al_2O_3}=0,1mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al_2O_3}=0,1\cdot102=10,2\left(g\right)\\m_{Mg}=0,1\cdot24=2,4\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{hh}=10,2+2,4=12,6\left(g\right)\)

Theo PTHH: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{Mg}=0,1mol\\n_{AlCl_3}=2n_{Al_2O_3}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\end{matrix}\right.\)

Mặt khác: \(\left\{{}\begin{matrix}m_{ddHCl}=400\cdot1,2=480\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=492,4\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{9,5}{492,4}\cdot100\%\approx1,93\%\\C\%_{AlCl_3}=\dfrac{26,7}{492,4}\cdot100\%\approx5,42\%\end{matrix}\right.\)

\(n_{KOH}=0.4\cdot2=0.8\left(mol\right)\)

\(2KOH+H_2SO_4\rightarrow K_2SO_4+H_2O\)

\(0.8..............0.4............0.4\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.4}{0.5}=0.8\left(l\right)\)

\(C_{M_{K_2SO_4}}=\dfrac{0.4}{0.4+0.8}=0.33\left(M\right)\)

\(2KOH+H2SO4\rightarrow K2SO4+2H2O\)

\(n_{KOH}=0,4.2=0,8\left(mol\right)\)

Theo PT:

\(n_{H2SO4}=n_{K2SO4}=\dfrac{1}{2}n_{KOH}=0,4\left(mol\right)\)

\(\Rightarrow V_{H2SO4}=\dfrac{0,4}{0,5}=0,8\left(l\right)\)

\(V_{dd}saupư_{K2SO4}=0,4+0,8=1,2\left(l\right)\)

\(C_{M\left(K2SO4\right)}=\dfrac{0,4}{1,2}=\dfrac{1}{3}\left(M\right)\)