\(n_{NaOH}=\dfrac{12,4}{62}=0,2\left(mol\right)\)

PTHH: Na₂O + H₂O --> 2NaOH
______0,2-------------->0,4

=> m_NaOH = 0,4.40 = 16(g)

=> \(C\%=\dfrac{16}{12,4+187,6}.100\%=8\%\)

=> B

46

\(m_{NaOH}=100\cdot4\%=4\left(g\right)\)

\(n_{NaOH}=\dfrac{4}{40}=0.1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(0.05.......................0.1\)

\(m_{Na_2O}=0.05\cdot62=3.1\left(g\right)\)

\(n_{Na_2O}=\dfrac{6.2}{62}=0.1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(0.1.........................0.2\)

\(C_{M_{NaOH}}=\dfrac{0.2}{0.5}=0.4\left(M\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

\(0.2.............0.2\)

\(m_{HCl}=0.2\cdot36.5=7.3\left(g\right)\)

Chúc em học tốt !!!

Hai bạn làm sai một số chỗ, mình sẽ làm lại

Bài 1:

\(Na_2O\left(0,1\right)+H_2O--->2NaOH\left(0,2\right)\)

\(n_{Na_2O}=0,1\left(mol\right)\)

\(\Rightarrow m_{NaOH}=0,2.40=8\left(g\right)\)

\(m_{ddsau}=6,2+73,8=80\left(g\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{8}{80}.100=10\%\)

Bài 2:

\(n_{Na_2O}=0,1\left(mol\right)\)

\(m_{NaOH}\left(bđ\right)=60\left(g\right)\)

\(\Rightarrow m_{H_2O}=133,8-60=73,8\left(g\right)\)\(\Rightarrow n_{H_2O}=4,1\left(mol\right)\)

\(Na_2O\left(0,1\right)+H_2O\left(0,1\right)--->2NaOH\left(0,2\right)\)

So sánh: \(\dfrac{n_{Na_2O}}{1}=0,1< \dfrac{n_{H_2O}}{1}=4,1\)

=> Chọn số mol của Na2O để tính

Theo PTHH: nNaOH (tạo thành) = 0,2 (mol)

=> mNaOH (tạo thành) = 8 (g)

\(\Rightarrow\sum m_{NaOH}\left(sau\right)=60+8=68\left(g\right)\)

\(m_{ddsau}=6,2+133,8=140\left(g\right)\)

\(\Rightarrow C\%_{NaOH}\left(sau\right)=\dfrac{68}{140}.100=48,57\%\)

Bài 3:

\(m_{NaOH}\left(bđ\right)=12\left(g\right)\)

\(n_{Na_2O}=\dfrac{a}{62}\left(mol\right)\)

\(Na_2O\left(\dfrac{a}{62}\right)+H_2O--->2NaOH\left(\dfrac{a}{31}\right)\)

\(m_{NaOH}\left(tao.thanh\right)=\dfrac{a}{31}.40=\dfrac{40a}{31}\left(g\right)\)

\(\Rightarrow\sum m_{NAoh}\left(sau\right)=12+\dfrac{40a}{31}\left(g\right)\)

\(m_{ddsau}=\left(a+120\right)\left(g\right)\)

Ta có: \(20=\dfrac{12+\dfrac{40a}{31}}{a+120}.100\)

\(\Rightarrow a=11\left(g\right)\)

BT 1:

m_dd = m_ct + m_dm = 6,2 + 73,8 = 80 (g)

C%_A = \(\dfrac{m_{ct}}{m_{dd}}.100=\dfrac{6,2}{80}.100=7,75\%\)

Bài tập 1:

m_dd = 6,2 + 73,8 = 80(g)

C% = \(\dfrac{6,2}{80}.100=7,75\%\)

PT : Na₂O + H₂O \(\rightarrow\) 2NaOH

7,75%------------ 15,5%

Vậy dung dịch A có nồng độ là 15,5%.

m dd- 4+46=50g

=>C% NaOH=4\50 .100=8%

=>A

\(n_{Na_2O}=\dfrac{6,2}{62}=0,1(mol)\\ PTHH:Na_2O+H_2O\to 2NaOH\\ \Rightarrow n_{NaOH}=0,2(mol)\\ \Rightarrow m_{CT_{NaOH}}=0,2.40=8(g)\\ \Rightarrow C\%_{NaOH}=\dfrac{8}{6,2+193,8}.100\%=4\%\)

\(n_{Na_2O}=\dfrac{m}{M}=\dfrac{6,2}{39}=0,1\) ( mol )

Na₂O + H₂O → 2NaOH

  0,1             →     0,2                ( mol )

⟹ m_NaOH = n_NaOH. M_NaOH = 0,2. 40 = 8 (g)

Khối lượng dung dịch sau là:

m_{dd sau} = \(m_{Na_2O}+m_{H_2O}\) = 6,2 + 193,8 = 200 (g)

dung dịch A thu được là dung dịch NaOH

⇒ C%_NaOH = ( m_NaOH : m _{dd sau pư} ) . 100% =( 8 : 200 ) .100% = 4%

Do mik  bấm phân số không được nên mik để phép chia nha

\(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\\ PTHH:Na_2O+H_2O\rightarrow2NaOH\\ \Rightarrow n_{NaOH}=0,2\left(mol\right)\\ \Rightarrow m_{CT_{NaOH}}=0,2\cdot40=8\left(g\right)\\ m_{dd_{NaOH}}=6,2+193,8=200\left(g\right)\\ \Rightarrow C\%_{NaOH}=\dfrac{8}{200}\cdot100\%=4\%\)

Gọi kl kiềm là A. 

A + H2O -> AOH + 1/2 2

nH2= 0,1(mol) -> nA=0,2(mol)

=>M(A)=mA/nA=4,6/0,2=23(g/mol)

->A là Natri (Na=23)

PTHH: Na + H2O -> NaOH + 1/2 H2

0,2_____________0,2________0,1(mol)

mNaOH=0,2.40=8(g)

mddNaOH=mNa+ mH2O - mH2= 4,6+95,6-0,1.2=100(g)

-> C%ddNaOH= (8/100).100=8%