1,dung môi là nước dung dịch là nước muối

2,ko bt

\(n_{K_2O}=\dfrac{6,58}{94}=0,07\left(mol\right);n_K=\dfrac{5,85}{39}=0,15\left(mol\right)\)

PTHH: \(4K+O_2\xrightarrow[]{t^o}2K_2O\)

            0,14<--------0,07

\(\Rightarrow H=\dfrac{0,14}{0,15}.100\%=93,33\%\)

PTHH: \(K_2O+H_2O\rightarrow2KOH\)

             0,07------------->0,14

\(\Rightarrow C_{M\left(KOH\right)}=\dfrac{0,14}{0,2}=0,7M\)

2R+2H2O->2ROH+H2

ta có :

\(\dfrac{12,4}{R}=\dfrac{16}{R+16+1}\)

=>R=58,55 

=>? chất

ta có

R2O+H2O->2ROH

ta có :

\(\dfrac{12,4}{x.2+16}=\dfrac{16}{2\left(x+16+1\right)}\)

=>R =23 (Na)

Giả sử oxit kim loại cần tìm là A₂O.

PT: \(A_2O+H_2O\rightarrow2AOH\)

Ta có: \(n_{A_2O}=\dfrac{3,1}{2M_A+16}\left(mol\right)\)

\(n_{AOH}=\dfrac{4}{M_A+17}\left(mol\right)\)

Theo PT: \(n_{AOH}=2n_{A_2O}\Rightarrow\dfrac{4}{M_A+17}=\dfrac{3,1.2}{2M_A+16}\)

\(\Rightarrow M_A=23\left(g/mol\right)\)

⇒ A là Na.

Vậy: Oxit đó là Na₂O.

Bạn tham khảo nhé!

a, Giả sử R có hóa trị n.

PT: \(R_2O_n+nH_2O\rightarrow2R\left(OH\right)_n\)

Theo ĐLBT KL, có: m _oxit + m_H2O = m _hydroxit

⇒ 3,1 + 18n_H2O = 4 ⇒ n_H2O = 0,05 (mol)

Theo PT: \(n_{R_2O_n}=\dfrac{1}{n}n_{H_2O}=\dfrac{0,05}{n}\left(mol\right)\)

\(\Rightarrow M_{R_2O_n}=\dfrac{3,1}{\dfrac{0,05}{n}}=62n\) \(\Rightarrow2M_R+16n=62n\Rightarrow M_R=23n\)

Với n = 1 thì M_R = 23 (g/mol)

→ R là Natri. Na₂O: natri oxit. NaOH: natri hydroxit.

b, PT: \(4Na+O_2\underrightarrow{t^o}2Na_2O\)

Ta có: \(n_{Na_2O}=\dfrac{3,1}{62}=0,05\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Na_2O}=0,025\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,025.22,4=0,56\left(l\right)\)

Gọi oxit kim loại cần tìm là RO

\(n_{RO}=\dfrac{5,6}{M_R+16}\left(mol\right)\)

PTHH: RO + H₂O --> R(OH)₂ 

         \(\dfrac{5,6}{M_R+16}\)---------->\(\dfrac{5,6}{M_R+16}\)

=> \(M_{R\left(OH\right)_2}=M_R+34=\dfrac{7,4}{\dfrac{5,6}{M_R+16}}=\dfrac{37}{28}\left(M_R+16\right)\left(g/mol\right)\)

=> M_R = 40 (g/mol)

=> Kim loại là Ca

\(Đặt.oxit:AO\\ AO+H_2O\rightarrow A\left(OH\right)_2\\Ta.có:m_{H_2O}=m_{bazo}-m_{oxit}=7,4-5,6=1,8\left(g\right)\\ n_{H_2O}=n_{AO}=\dfrac{1,8}{18}=0,1\left(mol\right)\\ \Rightarrow M_{AO}=\dfrac{5,6}{0,1}=56\left(\dfrac{g}{mol}\right)=M_A+16\\ \Leftrightarrow M_A=40\left(\dfrac{g}{mol}\right)\\ \Rightarrow A:Canxi\left(Ca=40\right)\)

1)

$n_{Na_2O} = \dfrac{6,2}{62} = 0,1(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,2(mol)$
$m_{dd} = 6,2 + 193,8 = 200(gam) \Rightarrow C\%_{NaOH} = \dfrac{0,2.40}{200}.100\% = 4\%$

2) 

$n_{K_2O} = \dfrac{23,5}{94} = 0,25(mol)$
$K_2O + H_2O \to 2KOH$
$n_{KOH} = 2n_{K_2O} = 0,5(mol) \Rightarrow C_{M_{KOH}} = \dfrac{0,5}{0,5} = 1M$

3) $n_{Na_2O} = \dfrac{12,4}{62} = 0,2(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,4(mol)$
$C_{M_{NaOH}} = \dfrac{0,4}{0,5} =0,8M$

4)

$Na_2SO_3 + 2HCl \to 2NaCl +S O_2 + H_2O$
Theo PTHH : 

$n_{SO_2} = n_{Na_2SO_3} = \dfrac{12,6}{126} = 0,1(mol)$
$V_{SO_2} = 0,1.22,4 = 2,24(lít)$

5) $n_{CaO} = \dfrac{5,6}{56} = 0,1(mol)$

$CaO + 2HCl \to CaCl_2 + H_2O$

Theo PTHH : 

$n_{HCl} = 2n_{CaO} = 0,2(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$

Bài 1 : 

\(n_{Na2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

Pt : \(Na_2O+H_2O\rightarrow2NaOH|\)

           1          1              2

          0,1                      0,2

\(n_{NaOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

⇒ \(m_{NaOH}=0,2.40=8\left(g\right)\)

\(m_{ddspu}=6,2+193,8=200\left(g\right)\)

\(C_{ddNaOH}=\dfrac{8.100}{200}=4\)⁰/₀

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