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

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

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%

a, \(n_{KOH}=\dfrac{44,8.25\%}{56}=0,2\left(mol\right);n_{H_2SO_4}=0,1.1,5=0,15\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ → K₂SO₄ + H₂O

Mol:       0,2         0,1              0,1

Ta có: \(\dfrac{0,2}{2}< \dfrac{0,15}{1}\) ⇒ KOH hết, H₂SO₄ dư

 ⇒ Khi cho quỳ tím vào sẽ lm quỳ tím chuyển đỏ (vì trong dd vẫn còn axit)

b) \(m_{ddH_2SO_4}=1,1.100=110\left(g\right)\)

⇒ m_{dd sau pứ} = 44,8 + 110 = 154,8 (g)

\(C\%_{ddK_2SO_4}=\dfrac{0,1.174.100\%}{154,8}=11,24\%\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 2K + 2H₂O --> 2KOH + H₂

            0,3<-------------0,3<---0,15

=> m_K = 0,3.39 = 11,7 (g)

=> m_KOH(A) = 21,1 - 11,7 = 9,4 (g)

m_{KOH(dd sau pư)} = 0,3.56 + 9,4 = 26,2 (g)

a = 200 + 0,15.2 - 21,1 = 179,2 (g)

\(C\%=\dfrac{26,2}{200}.100\%=13,1\%\) => x = 13,1 

Đáp án B

m C t   =   m K O H  = 1.56 = 56 gam.

m d d   =   m K L   +   m n ư ớ c   -   m k h í  = 39 + 362 – 0,5.1 = 400 gam.

Chọn B

\(CT:M\left(OH\right)_n\)

\(M\left(OH\right)_n+nHCl\rightarrow MCl_n+nH_2O\)

\(M+17n...........M+35.5n\)

\(7.4..........................11.37\)

\(\Leftrightarrow11.37\left(M+17n\right)=7.4\left(M+35.5n\right)\)

\(\Leftrightarrow3.97M-69.41n=0\)

\(\Leftrightarrow M=17.48n\)

\(n=3\Rightarrow M=52\)

\(CT:Cr\left(crom\right)\)