n\(_{Na}\)= \(\dfrac{6,9}{23}\)= 0,3 (mol)

m\(_{NaOH}\) = 150 . \(\dfrac{20}{100}\) = 30 (g)

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

mol: ....0,3------------------>0,3

\(\sum\)m\(_{NaOH}\) = 30 + 0,3 . 40 = 42 (g)

mdd\(_{NaOH}\) = 150 + 6,9 = 156,9 (g)

C% NaOH = \(\dfrac{42}{156,9}\).100% = 26,77%

a, \(C\%_{KCl}=\dfrac{20}{20+60}.100\%=25\%\)

b, \(C\%=\dfrac{40}{40+150}.100\%\approx21,05\%\)

c, \(C\%_{NaOH}=\dfrac{60}{60+240}.100\%=20\%\)

d, \(C\%_{NaNO_3}=\dfrac{30}{30+90}.100\%=25\%\)

e, \(m_{NaCl}=150.60\%=90\left(g\right)\)

f, \(m_{ddA}=\dfrac{25}{10\%}=250\left(g\right)\)

g, \(n_{NaOH}=120.20\%=24\left(g\right)\)

Gọi: n_{NaOH (thêm vào)} = a (g)

\(\Rightarrow\dfrac{a+24}{a+120}.100\%=25\%\Rightarrow a=8\left(g\right)\)

a, \(2Na+2H_2O\rightarrow2NaOH+H_2\)

Ta có: \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)

\(n_{H_2O}=\dfrac{200}{18}=\dfrac{100}{9}\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{\dfrac{100}{9}}{2}\), ta được H₂O dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)

b, Theo PT: \(n_{NaOH}=n_{Na}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 2,3 + 100 - 0,05.2 = 102,2 (g)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,1.40}{102,2}.100\%\approx3,91\%\)

c, - Dung dịch làm quỳ tím hóa xanh.

\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{H_2}=\dfrac{0,1}{2}=0,05\left(mol\right);n_{NaOH}=n_{Na}=0,1\left(mol\right)\\ a,V=V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ b,m_{ddNaOH}=m_{Na}+m_{H_2O}-m_{H_2}=2,3+200-0,05.2=202,2\left(g\right)\\ C\%_{ddNaOH}=\dfrac{40.0,1}{202,2}.100\approx1,978\%\\ c,NaOH-Tính.bazo\Rightarrow Quỳ.tím.hoá.xanh\)

Na(r) + 2H2O(l) 2 ------> NaOH(dd) + H2(k) 

 nNa = \(\dfrac{2,875}{23}\) = 0,125 (mol)

Theo PTHH: n_H2= \(\dfrac{1}{2}n_{Na}\).= 0,5. 0,125 = 0,0625 (mol)

nNaOH = nNa = 0,125 (mol)

Khối lượng của dung dịch thu được sau khi phản ứng kết thúc :

m_{dd NaOH} = 2,875 + 497,25 – 0,0625.2 = 500 (gam)

Khối lượng NaOH có trong dd thu được:

m_NaOH = \(500.10,945\%\)= 54,725 (gam)

Khối lượng NaOH có trong dd ban đầu là:

m_NaOH = 54,725g – 0,125. 40 = 49,725(gam)

Nồng độ % của dd NaOH ban đầu là: a = \(\dfrac{49,725}{497,25}.100\) = 10% 

\(C\%_{ddNaOH\left(thu.được\right)}=\dfrac{20}{20+150}.100\%\approx11,765\%\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

            \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)

Ta có: \(n_{Fe\left(OH\right)_2}=\dfrac{18}{90}=0,2\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,4\left(mol\right)\\n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\\C\%_{NaOH}=\dfrac{0,4\cdot40}{160}\cdot100\%=10\%\end{matrix}\right.\)