n_Na = \(\dfrac{m}{M}=\dfrac{6,9}{23}\) = 0,3(mol)

Gọi x là khối lượng của NaOH 80%

PTHH: 2Na + 2H₂O → 2NaOH + H2 (1)

TPT: 2mol 2mol 1mol

TĐB: 0,3mol ?(mol) ?(mol)

=>n_H2 = \(\dfrac{0,3.1}{2}\)= 0,15(mol); n_{NaOH (1)} = \(\dfrac{0,3.2}{3}\) = 0,3(mol)

=> m_H2 = n . M = 0,15 . 2 = 0,3(g)

=> m_NaOH(1) = n . M = 0,3 . 40 = 12(g)

n_Na2O = \(\dfrac{m}{M}=\dfrac{9,3}{\left(23.2+16\right)}\) = 0,15(mol)

PTHH: Na₂O + H₂O → 2NaOH (2)

TPT: 1mol 2mol

TĐB: 0,15(mol) ?(mol)

=> n_{NaOH (2)} = \(\dfrac{0,15.2}{1}\) = 0,3(mol)

=> m_NaOH(2) = n . M = 0,3 . 40 = 12(g) \

=> Tổng m_NaOH(8%) = m_{NaOH (1)} + m_{NaOH (2)} = 12g +12g = 24g

=> m_ddNaOH(8%) = \(\dfrac{m_{ctan}}{C\%}.100\%\) = \(\dfrac{24}{8\%}.100\%\) = 300(g)
Gọi m_{dd NaOH} cần thêm vào là: a(g)
=> m_NaOH = 80% . a + 24
=> m_dd = 300 + a
Ta có: (80% . a + 24) : (300 + a) = 15%
=> a \(\approx\) 32,31(g)

Vậy chúng ta cần phải lấy thêm 32,31g dd NaOH có độ tinh khiết 80% cho vào dung dịch A để được dd 15%

PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\)

            \(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=n_{Na}+2n_{Na_2O}=\dfrac{4,6}{23}+2\cdot\dfrac{6,2}{62}=0,3\left(mol\right)\\n_{H_2}=\dfrac{1}{2}n_{Na}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,3\cdot40=12\left(g\right)\\m_{H_2}=0,05\cdot2=0,1\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{Na}+m_{Na_2O}+m_{H_2O}-m_{H_2}=110,7\left(g\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{12}{110,7}\cdot100\%\approx10,84\%\)

Theo gt ta có: $n_{Na}=0,2(mol);n_{Na_2O}=0,1(mol)$

$2Na+2H_2O\rightarrow 2NaOH+H_2$

$Na_2O+H_2O\rightarrow 2NaOH$

Ta có: $n_{NaOH}=0,4(mol);n_{H_2}=0,1(mol)$

Bảo toàn khối lượng ta có: $m_{dd}=110,6(g)$

$\Rightarrow \%C_{NaOH}=14,46\%$

\(n_{Na}=\dfrac{m}{M}=0,2mol\)

\(n_{Na2O}=\dfrac{m}{M}=0,1\left(mol\right)\)

\(BTNT\left(Na\right):n_{NaOH}=0,2+2.0,1=0,4\left(mol\right)\)

\(\Rightarrow m_{NaOH}=n.M=16g\)

\(PTHH:Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

...............0,2...................................0,1.

\(\Rightarrow m_{dd}=m_{KL}+m_{H2O}-m_{H2}=110,6g\)

\(\Rightarrow C\%_{NaOH}=\dfrac{m}{m_{dd}}.100\%=14,46\%\)

Coi X gồm Na, Ba và O.

Ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{20,52}{171}=0,12\left(mol\right)\)

BTNT Ba, có: n_Ba = n_Ba(OH)2 = 0,12 (mol)

Gọi: \(\left\{{}\begin{matrix}n_{Na}=x\left(mol\right)\\n_O=y\left(mol\right)\end{matrix}\right.\) ⇒ 23x + 16y = 21,9 - 0,12.137 (1)

Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

BT e, có: n_Na + 2n_Ba = 2n_O + 2n_H2 ⇒ x + 0,12.2 = 2y + 0,05.2 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,14\left(mol\right)\\y=0,14\left(mol\right)\end{matrix}\right.\)

BTNT Na, có: n_NaOH = n_Na = 0,14 (mol) 

⇒ m = m_NaOH = 0,14.40 = 5,6 (g)

Ta có \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(m_{NaOH}=100.16\%=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)

PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(Na_2O+H_2O\rightarrow2NaOH\)

Theo PT: \(n_{Na}=2n_{H_2}=0,2\left(mol\right)\)

\(n_{NaOH}=n_{Na}+2n_{Na_2O}\Rightarrow n_{Na_2O}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,2.23}{0,2.23+0,1.62}.100\%\approx42,6\%\\\%m_{Na_2O}\approx57,4\%\end{matrix}\right.\)

\(n_{Na}=\dfrac{13,8}{23}=0,6\left(mol\right)\\ pthh:Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\) 
          0,6                    0,6                  0,3 
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\\ c,m_{\text{dd}}=13,8+286,8-\left(0,3.2\right)=300\left(g\right)\\ C\%=\dfrac{0,6.40}{300}.100\%=8\%\)

\(n_{Na}\) = \(\dfrac{13,8}{23}\) = 0,6 mol

Theo PTHH:

a) \(2Na+2H_2O\underrightarrow{t^o}2NaOH+H_2\)

     2          2           2                1 (mol)

    0,6 \(\rightarrow\) 0,6   \(\rightarrow\) 0,6     \(\rightarrow\)    0,3 (mol)

b) \(V_{H_2}\) = 0,3.22,4 = 6,72l

c) \(m_{dd}\) = 13,8 + 286,8  - 0,3.2 = 300g

\(C\%\) = \(\dfrac{0,6.40}{300}\).100% = 8%

2. Ta có: m_{KOH 20%} = \(\frac{200.20}{100}\) ₌ 40g

m_{KOH 10%} = m_KOH 20% = 40g

=>m_{dd KOH 10%}  = \(\frac{40.100}{10}\) = 400g

=> m_H2O = 400 - 200 =200g

80 gam dung dịch A chứa 3,52 gam NaOH

=> 200 gam dung dịch A chứa 3,52.200/80 = 8,8 gam

n NaOH = 8,8/40 = 0,22(mol)

Gọi n Na = a(mol) ; n Na2O = b(mol)

=> 23a + 62b = 6,02(1)

$2Na + 2H_2O \to 2NaOH + H_2$
$Na_2O + H_2O \to 2NaOH$
n NaOH = a + 2b = 0,22(2)

Từ (1)(2) suy ra  a= 0,1 ; b = 0,06

n H2 = 0,5a = 0,05(mol)

=> m H2O = 200 + 0,05.2 - 6,02 =194,08(gam)

%m Na = 0,1.23/6,02   .100% = 38,2%

%m Na2O = 100% -38,2% = 61,8%