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

b)dd B là H₂SO₄ dư

số mol h2so4 là:

n_H2SO4(dư) = 0,1 - 0,02 = 0,08 mol

Nồng độ phần trăm dd B là :

C%_H2SO4 =\(\dfrac{0,08.98}{50+200-4,66}.100\%\)3,2 %

a) \(n_{H_2SO_4}=\dfrac{50.19,6}{100.98}=0,1\left(mol\right)\)

\(n_{Ba\left(OH\right)_2}=\dfrac{200.17,1}{100.171}=0,2\left(mol\right)\)

PTHH: Ba(OH)₂ + H₂SO₄ --> BaSO₄ + 2H₂O

Xét \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\) => Ba(OH)₂ dư, H₂SO₄ hết

PTHH: Ba(OH)₂ + H₂SO₄ --> BaSO₄ + 2H₂O

_______0,1<-----0,1--------->0,1

=> m_BaSO4 = 0,1.233 = 23,3 (g) 

b) m_{dd sau pư} = 50 + 200 - 23,3 = 226,7 (g)

=> \(C\%=\dfrac{\left(0,2-0,1\right).171}{226,7}.100\%=7,543\%\)

\(n_{BaCl_2}=\dfrac{200.10,4\%}{208}=0,1\left(mol\right)\\ a,BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ b,Qu\text{ỳ}-t\text{í}m-ho\text{á}-\text{đ}\text{ỏ}-do-c\text{ó}-\text{ax}it-HCl\\ c,n_{BaSO_4}=n_{H_2SO_4}=n_{BaCl_2}=0,1\left(mol\right)\\ m_{k\text{ết}-t\text{ủa}}=m_{BaSO_4}=233.0,1=23,3\left(g\right)\\ d,m_{\text{dd}H_2SO_4}=\dfrac{0,1.98}{4,9\%}=200\left(g\right)\\ e,m_{\text{dd}HCl}=200+200-23,3=376,7\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ C\%_{\text{dd}HCl}=\dfrac{0,2.36,5}{376,7}.100\approx1,938\%\)

Bài 19  :

\(a) n_{Al} = \dfrac{10,8}{27} = 0,4(mol)\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,6(mol)\\ V_{H_2} = 0,6.22,4 = 13,44(lít)\\ b) \text{Chất tan : }Al_2(SO_4)_3\\ n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,2(mol)\\ m_{Al_2(SO_4)_3} = 0,2.342 = 68,4(gam)\)

Bài 18 : 

\(a) n_{HCl} = \dfrac{250.7,3\%}{36,5 } = 0,5(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,25(mol) \Rightarrow V_{H_2} = 0,25.22,4 = 5,6(lít)\\ b) \text{Chất tan : } ZnCl_2\\ n_{ZnCl_2} = n_{H_2} = 0,25(mol)\\ m_{ZnCl_2} = 0,25.136 = 34(gam)\)

\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)

0,2                  0,075       0,075      0,15    (mol)

\(m\) kết tủa = 0,075.233=17,475

C%=\(\dfrac{17,475}{250+150-0,15.18}.100\approx4,398\)

Mik làm rồi nhé

Bn vào link này: https://hoc24.vn/cau-hoi/cho-250g-dung-dich-baoh2-342-tac-dung-voi-150-gam-dung-dich-h2so4-49-tinh-khoi-luong-ket-tua-thu-duoccho-250-g-dung-dich-baoh2-342-tac-dung-voi-150-gam-dung-dich-h2so4-49a-tinh-khoi-luon.2472774496310

\(n_{H_2SO_4}=\dfrac{200.19,6}{100.98}=0,4mol\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4\left(A\right)}=n_{CuO}=n_{H_2SO_4}=0,4mol\\ n_{Cu\left(OH\right)_2}=\dfrac{29,4}{98}=0,3mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\\Rightarrow\dfrac{0,4}{1}>\dfrac{0,3}{1}\Rightarrow CuSO_4.pư.không.hết\)

\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

0,3mol        0,6mol                                        0,3mol

\(m_{ddB}=0,4.80+200+0,6.40-29,4=226,6g\\ C_{\%Na_2SO_4\left(B\right)}=\dfrac{0,3.142}{226,6}\cdot100=18,8\%\)