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Câu 4 :
\(n_{SO2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Pt : \(SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3+H_2O|\)
1 1 1 1
0,15 0,15 0,15
a) \(n_{Ba\left(OH\right)2}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
⇒ \(m_{Ba\left(OH\right)2}=0,15.171=25,65\left(g\right)\)
\(C_{ddBa\left(OH\right)2}=\dfrac{25,65.100}{150}=17,1\)0/0
b) \(n_{BaSO3}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
⇒ \(m_{BaSO3}=0,15.217=32,55\left(g\right)\)
c) Pt : \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O|\)
1 2 1 2
0,15 0,3
\(n_{HCl}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)
\(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
\(m_{ddHCl}=\dfrac{10,95.100}{20}=54,75\left(g\right)\)
\(V_{ddHCl}=\dfrac{54,75}{1,2}=45,625\left(ml\right)\)
Chúc bạn học tốt
\(a.n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\\ 0,1...........0,1.............0,1..........0,1\left(mol\right)\\ b.m_{kt}=m_{BaCO_3}=0,1.197=19,7\left(g\right)\\ c.C_{MddBa\left(OH\right)_2}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
a, \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: CO2 + Ca(OH)2 → CaCO3 ↓ + H2O
Mol: 0,25 0,25 0,25
\(C_{M_{ddCa\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5M\)
b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)
c,
PTHH: Ca(OH)2 + 2HCl → CaCl2 + 2H2O
Mol: 0,25 0,5
\(m_{ddHCl}=\dfrac{0,5.36,5.100}{20}=91,25\left(g\right)\)
a, \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
b, \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{Ca\left(OH\right)_2}=n_{CO_2}=0,25\left(mol\right)\Rightarrow C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,2}=1,25\left(M\right)\)
c, \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)
Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{15\%}\approx121,67\left(g\right)\)
\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\\ a)CO_2+Ca\left(OH\right)_2\rightarrow Ca\left(OH\right)_2+H_2O\)
0,25 0,25 0,25
\(b)C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,2}=1,25M\\ c)2HCl+Ca\left(OH\right)_2\rightarrow CaCl_2+2H_2O\\ n_{HCl}=2n_{Ca\left(OH\right)_2}=2.0,25=0,5mol\\ m_{ddHCl}=\dfrac{0,5.36,5}{15\%}\cdot100\%\approx121,67g\)
PTHH : `Ba(OH)_2 + SO_2 -> BaSO_3 + H_2O`
`a)`
`600ml = 0,6l`
`n_{SO_2} = (6,72)/(22,4) = 0,3` `mol`
`n_{Ba(OH)_2} = n_{SO_2} = 0,3` `mol`
`C_{M_(Ba(OH)_2)} = (0,3)/(0,6) =0,5` `M`
`b)`
`n_{BaSO_3} = n_{SO_3} = 0,3` `mol`
`m_{BaSO_3} = 0,3 . 217 = 65,1` `gam`
`c)`
PTHH : `Ba(OH)_2 + 2HCl -> BaCl_2 + 2H_2O`
Ta có : `n_{Ba(OH)_2} = 0,3` `mol`
`n_{HCl} = 2 . n_{Ba(OH)_2} = 0,6` `mol`
`V_{HCl} = (0,6)/(3,5) = 6/35` `l`