\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\\ 200ml=0,2l\\ C_{M_{NaOH}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

Giả sử có 100g dd kiềm như đề bài thì ta có

31g NaOH và 69g H₂O

=> n_NaOH = \(\frac{31}{40}\)= 0,775 (mol)

Ta có PT:

Na₂O + H₂O ---> 2NaOH

0,3875.......0,3875 ........0,775

=> m_{\(Na_2O\)} = 62.0,3875 = 24,025(g)

m_{\(H_2O\)} = 100 - 24,025 = 75,975(g)

vậy hòa tan 24,025 g Na₂O và 75,975g nước để tạo được dd kiềm 31%

\(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=0,2.2=0,4\left(mol\right)\\ C_{MddNaOH}=\dfrac{0,4}{0,4}=1\left(M\right)\)

\(n_{Na_2O}=\dfrac{12,4}{62}=0,2mol\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=0,2.2=0,4mol\\ C_{M_{NaOH}}=\dfrac{0,4}{0,4}=1M\)

\(n_{Na_2O}=\dfrac{12,4}{62}=0,2mol\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,2  \(\rightarrow\)    0,2   \(\rightarrow\)   0,4

\(C_{M_{NaOH}}=\dfrac{0,4}{\dfrac{500}{1000}}=0,8M\)

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

PTHH: Na₂O + H₂O ---> 2NaOH (1)

a. Theo PT₍₁₎: \(n_{NaOH}=2.n_{Na_2O}=2.0,05=0,1\left(mol\right)\)

=> \(C_{M_{NaOH}}=\dfrac{0,1}{2}=0,05M\)

b. PTHH: 2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}.n_{NaOH}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

=> \(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{4,9}{m_{dd_{H_2SO_4}}}.100\%=20\%\)

=> \(m_{dd_{H_2SO_4}}=24,5\left(g\right)\)

cau 1

chữ xấu quá

Ta có: \(C\%_{KOH}=\dfrac{m_{KOH}}{m_{KOH}+m_{H_2O}}.100\%\)

\(\Rightarrow\dfrac{m_{KOH}}{m_{KOH}+400}=0,12\)

\(\Rightarrow m_{KOH}=\dfrac{600}{11}\left(g\right)\)

\(a,PTHH:Na_2O+H_2O\rightarrow2NaOH\\ \Rightarrow n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{37,2}{62}=0,6\cdot2=1,2\left(mol\right)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{1,2}{0,5}=2,4M\\ b,PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,6\cdot98=58,8\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{58,8\cdot100\%}{20\%}=294\left(g\right)\\ \Rightarrow V_{dd}=\dfrac{294}{1,14}\approx257,9\left(ml\right)\)