\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH CO₂  + Ba(OH)₂ --> BaCO₃  + H₂O

CO₂ phản ứng với Ba(OH)₂ tạo muối trung hòa

n_Ba(OH)2 = n_CO2=0,1 mol

=> \(CM_{Ba\left(OH\right)_2}=\dfrac{0,1}{0,2}=0,5M\)

n_BaCO3 = n_CO2=0,1mol

=> \(m_{BaCO_3}=0,2.197=19,7\left(g\right)\)

\(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

a, \(n_{H_2}=n_{Zn}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)

b, \(n_{HCl}=2n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,3.36,5}{25\%}=43,8\left(g\right)\)

c, \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,15}{2,5}=0,06\left(l\right)\)

\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

a, \(n_{H_2}=n_{Mg}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)

b, \(n_{HCl}=2n_{Mg}=0,3\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,3.36,5}{25\%}=43,8\left(g\right)\)

c, PT: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,15}{2,5}=0,06\left(l\right)\)

n Hcl pu la 0,95*2 = 0,39 mol
n Hcl du la 0,5 -0,39 = 0,11 mol
gọi v lít là thể tích dung dịch kiềm
n Naoh la 0,2V mol,nBaoh la 0,1V mol
pthh .....bạn ghi ra 2 pthh giua naoh voi hcl,baoh vs hcl
Ta co 0,4V =0,11
suy ra V =0,275 L

\(a.n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\\ 0,1...........0,1.............0,1..........0,1\left(mol\right)\\ b.m_{kt}=m_{BaCO_3}=0,1.197=19,7\left(g\right)\\ c.C_{MddBa\left(OH\right)_2}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

Dd HCl có CM không bạn?

\(m_{Ba\left(OH\right)_2}=\dfrac{100\cdot17,1\%}{100\%}=17,1\left(g\right)\\ n_{Ba\left(OH\right)_2}=\dfrac{17,1}{171}=0,1\left(mol\right)\\ PTHH:Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+H_2O\\ n_{BaCl_2}=n_{Ba\left(OH\right)_2}=0,1\left(mol\right)\\ \Rightarrow m_{muối.sau.p/ứ}=m_{BaCl_2}=0,1\cdot208=20,8\left(g\right)\)

n Hcl pu la 0,95*2 = 0,39 mol
n Hcl du la 0,5 -0,39 = 0,11 mol
gọi v lít là thể tích dung dịch kiềm
n Naoh la 0,2V mol,nBaoh la 0,1V mol
pthh .....bạn ghi ra 2 pthh giua naoh voi hcl,baoh vs hcl
Ta co 0,4V =0,11
suy ra V =0,275 L