\(BaO+H_2SO_4->BaSO_4+H_2O\\ BaO+H_2O->Ba\left(OH\right)_2\\ n_{BaO}=\dfrac{30,6}{153}=0,2mol\\ n_{H_2SO_4}=0,098\cdot\dfrac{50}{98}=0,05mol\\ BaOdư\left(0,15mol=n_{Ba\left(OH\right)_2}\right)\\ C\%_{Ba\left(OH\right)_2}=\dfrac{0,15.171}{30,6+50-233.0,05}.100\%=37,2\%\)

\(BaO+H_2SO_4->BaSO_4+H_2O\\ n_{BaO}=\dfrac{30,6}{153}=0,2mol\\ n_{H_2SO_4}=0,098\cdot\dfrac{50}{98}=0,05mol\)

Vì acid hết, BaO dư nên C% dung dịch sau bằng 0%

a) \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

Theoo PTHH: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b) \(m_{dd\left(sau\right)}=250+6,5-0,1.2=256,3\left(g\right)\)

\(\Rightarrow C\%=\dfrac{0.1.161}{256,3}.100\%\approx0,06\%\)

c) \(V_{dd}=m_{dd}.D=256,3.1,15\approx294\left(ml\right)=0,294\left(l\right)\)

\(\Rightarrow C_M=\dfrac{0,1}{0,294}=0,34M\)

\(n_{NaOH}=1.0,5=0,5(mol)\\ 2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ \Rightarrow n_{H_2SO_4}=0,25(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,25.98}{9,8\%}=250(g)\)

Ta có: \(C_{\%_{Ba\left(OH\right)_2}}=\dfrac{m_{Ba\left(OH\right)_2}}{250}.100\%=34,2\%\)

=> \(m_{Ba\left(OH\right)_2}=85,5\left(g\right)\)

=> \(n_{Ba\left(OH\right)_2}=\dfrac{85,5}{171}=0,5\left(mol\right)\)

Ta lại có: \(C_{\%_{H_2SO_4}}=\dfrac{m_{H_2SO_4}}{150}.100\%=4,9\%\)

=> \(m_{H_2SO_4}=7,35\left(g\right)\)

=> \(n_{H_2SO_4}=\dfrac{7,35}{98}=0,075\left(mol\right)\)

a. PTHH; Ba(OH)₂ + H₂SO₄ ---> BaSO₄↓ + 2H₂O

Ta thấy: \(\dfrac{0,5}{1}>\dfrac{0,075}{1}\)

Vậy Ba(OH)₂ dư.

Theo PT: \(n_{BaSO_4}=n_{H_2SO_4}=0,075\left(mol\right)\)

=> \(m_{BaSO_4}=0,075.233=17,475\left(g\right)\)

b. Ta có: \(m_{dd_{BaSO_4}}=250+7,35=257,35\left(g\right)\)

=> \(C_{\%_{BaSO_4}}=\dfrac{17,475}{257,35}.100\%=6,79\%\)

Ko cần nhé

Chỉ hỏi bazơ thôi, ko dư nhé

\(n_{Fe2O3}=\dfrac{16}{160}=0,1\left(mol\right)\)

a) Pt : \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

            0,1-------->0,3------------>0,1

b) \(m_{ddH2SO4}=\dfrac{0,3.98}{9,8\%}.100\%=300\left(g\right)\)

c) \(m_{ddspu}=16+300=316\left(g\right)\)

\(C\%_{Fe2\left(SO4\right)3}=\dfrac{0,1.400}{316}.100\%=12,66\%\)

`Fe + H_2 SO_4 -> FeSO_4 + H_2`

`0,2`         `0,2`                `0,2`     `0,2`      `(mol)`

`n_[Fe] = [ 11,2 ] / 56 = 0,2 (mol)`

`a) V_[H_2] = 0,2 . 22,4 = 4,48 (l)`

`b) m_[dd HCl] = [ 0,2 . 98 ] / [ 9,8 ] . 100 = 200 (g)`

`c) C%_[FeSO_4] = [ 0,2 . 152 ] / [ 11,2 + 200 - 0,2 . 2 ] . 100 ~~ 14,42%`

\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)

Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)

         1           1                1            1

         0,3       0,3                            0,3

a) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

b) \(n_{H2SO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(m_{H2SO4}=0,3.98=29,4\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{29,4.100}{9,8}=300\left(g\right)\)

 Chúc bạn học tốt

\(a,n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

PTHH: Mg + 2HCl ---> MgCl₂ + H₂

LTL: \(0,2< \dfrac{0,5}{2}\) => HCl dư

Theo pthh: n_H2 = n_Mg = 0,2 (mol)

=> V_H2 = 0,2.22,4 = 4,48 (l)

\(b,n_{Fe}=\dfrac{2,8}{56}=0,.05\left(mol\right)\\ n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

PTHH: Fe + H₂SO₄ ---> FeSO₄ + H₂

LTL: 0,05 < 0,1 => H₂SO₄ dư

Theo pthh: n_H2 = n_Fe = 0,05 (mol)

=> V_H2 = 0,05.22,4 = 1,12 (l)

\(c,n_{Zn}=\dfrac{14,95}{65}=0,23\left(mol\right)\\ n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

LTL: \(0,23< \dfrac{0,6}{2}\) => HCl dư

Theo pthh: n_H2 = n_Zn = 0,23 (mol)

=> V_H2 = 0,23.22,4 = 5,152 (l)

a, 4,48 lít

b, 1,12 lít 

c, 5,512 lít

a, \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

b, Ta có: \(m_{H_2SO_4}=200.9,8\%=19,6\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)

Theo PT: \(n_{MgO}=n_{MgSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)

c, Ta có: m _{dd sau pư} = 8 + 200 = 208 (g)

\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,2.120}{208}.100\%\approx11,54\%\)