a) H₂SO₄ + 2NaOH --> Na₂SO₄ + 2H₂O

b) \(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)

PTHH: H₂SO₄ + 2NaOH --> Na₂SO₄ + 2H₂O

                 0,2---->0,4

=> m_NaOH = 0,4.40 = 16 (g)

=> \(m_{dd.NaOH}=\dfrac{16.100}{20}=80\left(g\right)\)

c)

PTHH: H₂SO₄ + 2KOH --> K₂SO₄ + 2H₂O

                0,2---->0,4

=> m_KOH = 0,4.56 = 22,4 (g)

=> \(m_{dd.KOH}=\dfrac{22,4.100}{5,6}=400\left(g\right)\)

=> \(V_{dd.KOH}=\dfrac{400}{1,045}=382,775\left(ml\right)\)

\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\\ pthh:H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\) 
          0,2                   0,4 
\(m_{\text{ }NaOH}=0,4.40=16g\\ m_{\text{dd}NaOH}=\dfrac{16.100}{20}=80g\) 
\(pthh:H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\) 
            0,2             0,4 
\(m_{KOH}=0,4.56=22,4g\\ m_{\text{dd}KOH}=\dfrac{22,4.100}{5,6}=400g\\ V_{\text{dd}}=\dfrac{400}{1,045}=382,7ml\)
\(V_{dd}=\dfrac{22,4}{1,045}=21,4354ml\)

\(n_{H2SO4}=0,02.1=0,02\left(mol\right)\)

PTHH: \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

pư..............0,02..........0,04..............0,02...........0,04 (mol)

\(\Rightarrow m_{NaOH}=0,04.40=1,6\left(g\right)\)

\(\Rightarrow m_{ddNaOH\left(20\%\right)}=\dfrac{1,6}{20\%}=8\left(g\right)\)

PTHH: \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)

pư............0,02............0,04............0,02..........0,04 (mol)

\(\Rightarrow m_{KOH}=0,04.56=2,24\left(g\right)\)

\(\Rightarrow m_{ddKOH\left(5,6\%\right)}=\dfrac{2,24}{5,6\%}=40\left(g\right)\)

\(\Rightarrow V_{KOH}=\dfrac{40}{1,045}\approx38,28\left(ml\right)\)

Vậy..............

,Em cảm ơn ạ <3

2

b

mNaCl=\(\dfrac{200.15}{100}\)=30(g)

nNaCl=\(\dfrac{30}{58,5}\)=0.51(mol)

VddNaCl=\(\dfrac{200}{1,1}\)=181.8(ml)=0.1818(l)

CMNaCl=\(\dfrac{0,51}{0,1818}\)=2.8(M)

dd la j bn

là dung dịch bạn

a. PTPỨ:  H₂SO₄ +  2NaOH \(\rightarrow\)  2H₂O +  Na₂SO₄

b. Ta có : n_H2SO4 = \(\frac{1.20}{1000}\) = 0,02 mol

c. Theo phương trình: n_NaOH = 2.n_H2SO4 = 2.0,02 = 0,04 mol

\(\Rightarrow\) m_NaOH = 0,04. 40 = 1,6(g)

d. m_{dd NaOH} = \(\frac{1,6.100}{20}\) = 8(g)

e₁.  PTHH: H₂SO₄ + 2KOH \(\rightarrow\)  K₂SO₄ + 2H₂O

Ta có: n_KOH = 2. n_H2SO4 = 2. 0,02 = 0,04 mol

\(\Rightarrow\) m_KOH = 0,04.56=2,24(g)

e₂. m_{dd KOH} = \(\frac{2,24.100}{5,6}\) = 40(g)

e₃. V_{dd KOH} = \(\frac{40}{1,045}\) \(\approx\) 38,278 ml

Ptrình ion H(+) + OH(-) = H2O 
n H(+) 0,3*0,75*2 + 0,3*1,5 = 0,9mol 
=> n OH(-) = 0,9mol => n KOH = 0,9mol => V = 0,6l

Cảm ơn bạn :)

\(n_{H_2SO_4}=0.3\cdot1=0.3\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

\(0.6...............0.3...............0.3\)

\(V_{dd_{NaOH}}=\dfrac{0.6}{1}=0.6\left(l\right)=600\left(ml\right)\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.3}{0.3+0.6}=0.33\left(M\right)\)

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

số mol H₂SO₄ :

n_H2SO4 = 0,5 . 0,7 = 0,35 (mol)

theo phương trình hóa học ta có:

n_KOH = 2.n_H2SO4 = 2.0,35 =0,7 (mol)

=> m_KOH = 0,7.56 = 39,2 (g)

=> m_{dd KOH} = (39,2.100)/12 =326,67 (g)

ta có :

V_{dd KOH} = 326,67/1,15 = 284 (ml)