nH2SO4=0,1(mol) 

a) PTHH: 2 NaOH + H2SO4 -> Na2SO4 + 2 H2O

b) 0,2___________0,1________0,1(mol)

mNaOH=0,2.40=8(g)

=>mddNaOH=(8.100)/25= 32(g)

c) mNa2SO4=0,1.142=14,2(g)

d) PTHH: 2 KOH + H2SO4 -> K2SO4 +2 H2O

nKOH=0,2(mol) => mKOH=0,2.56=11,2(g)

=> mddKOH=(11,2.100)/8=140(g)

=> VddKOH= 140/1,085=129,03(ml)

Chúc em học tốt!

thank you

PTHH: \(H_2SO_4+2KOH\rightarrow K_2SO+2H_2O\)

Ta có: \(n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{KOH}=0,4\left(mol\right)\\n_{K_2SO_4}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{ddKOH}=\dfrac{\dfrac{0,4\cdot56}{6\%}}{1,048}\approx356,2\left(ml\right)\\C_{M_{K_2SO_4}}=\dfrac{0,2}{0,2+0,3562}\approx0,36\left(M\right)\end{matrix}\right.\)   

\(n_{H_2SO_4}=1.0,2=0,2\left(mol\right)\\ H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\\ 0,2.........0,4........0,2.......0,2\left(mol\right)\\ a.m_{ddKOH}=\dfrac{0,4.56.100}{6}=\dfrac{1120}{3}\left(g\right)\\ V_{ddKOH}=\dfrac{\dfrac{1120}{3}}{1,048}=\dfrac{140000}{393}\left(ml\right)\approx0,356\left(l\right)\)

\(b.C_{MddK_2SO_4}=\dfrac{0,2}{\dfrac{140000}{393}+0,2}\approx0,00056\left(M\right)\)

a/ \(m_{H_2SO_4}=490.10\%=49\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol:         1             0,5             0,5

\(m_{ddNaOH}=\dfrac{1.40.100}{20}=200\left(g\right)\)

b/ \(m_{Na_2SO_4}=0,5.142=71\left(g\right)\)

Đổi 100ml = 0,1 lít

Ta có: \(C_{M_{HCl}}=\dfrac{n_{HCl}}{0,1}=1M\)

=> n_HCl = 0,1(mol)

PTHH: KOH + HCl ---> KCl + H₂O

Theo PT: \(n_{KOH}=n_{HCl}=0,1\left(mol\right)\)

=> \(m_{KOH}=0,1.56=5,6\left(g\right)\)

Ta có: \(C_{\%_{KOH}}=\dfrac{5,6}{m_{dd_{KOH}}}.100\%=20\%\)

=> \(m_{dd_{KOH}}=28\left(g\right)\)

thanks bro !!