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\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ PTHH:CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3\downarrow+H_2O\\ \Rightarrow n_{Ba\left(OH\right)_2}=n_{BaCO_3}=0,1\left(mol\right)\\ \Rightarrow C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,1}{0,2}=0,5M\\ m_{BaCO_3}=0,1\cdot197=19,7\left(g\right)\)
a/ \(n_{CO_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
PTHH: CO2 + Ba(OH)2 → BaCO3 + H2O
Mol: 0,3 0,3
b/ \(C_{M_{ddBa\left(OH\right)_2}}=\dfrac{0,3}{0,2}=1,5M\)
nCO2 =\(\dfrac{4,48}{22,4}\)=0,2 mol
PTHH CO2 + Ba(OH)2 --> BaCO3 + H2O
CO2 phản ứng vừa đủ với Ba(OH)2 => nBa(OH)2 = 0,2 mol
=> mBa(OH)2 = 0,2.171 = 34,2 gam
khối lượng 200ml dung dịch Ba(OH)2 có d = 1,12g/ml = 200.1,12 = 224 gam
C%Ba(OH)2 = \(\dfrac{m_{\left(ct\right)}}{m_{\left(dd\right)}}.100\)= \(\dfrac{34,2}{224}.100\)= 15,27%
2.
a, \(n_{HCl}=0,2.3,5=0,7\left(mol\right)\)
PTHH: CuO + 2HCl → CuCl2 + H2O
Mol: x 2x
PTHH: Fe2O3 + 6HCl → 2FeCl3 + 3H2O
Mol: y 6y
Ta có: \(\left\{{}\begin{matrix}80x+160y=20\\2x+6y=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
b, \(m_{CuO}=0,05.80=4\left(g\right);m_{Fe_2O_3}=20-4=16\left(g\right)\)
c,
PTHH: CuO + 2HCl → CuCl2 + H2O
Mol: 0,05 0,05
PTHH: Fe2O3 + 6HCl → 2FeCl3 + 3H2O
Mol: 0,1 0,2
\(m_{CuCl_2}=0,05.135=6,75\left(g\right)\)
\(m_{FeCl_3}=0,1.162,5=16,25\left(g\right)\)
1.
a, \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: CO2 + Ba(OH)2 → BaCO3 + H2O
Mol: 0,1 0,1 0,1
b, \(C_{M_{ddBa\left(OH\right)_2}}=\dfrac{0,1}{0,2}=0,5M\)
c, \(m_{BaCO_3}=0,1.197=19,7\left(g\right)\)
a) Ba(OH)2 + CO2 --> BaCO3 + H2O
b) \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Ba(OH)2 + CO2 --> BaCO3 + H2O
_0,1<------0,1------>0,1
=> mBaCO3 = 0,1.197 = 19,7 (g)
c) \(V_{dd\left(Ba\left(OH\right)_2\right)}=\dfrac{0,1}{0,5}=0,2\left(l\right)\)
\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1mol\\ CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\\ n_{BaCO_3}=n_{CO_2}=0,1mol\\ Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\\ n_{H_2SO_4}=n_{Ba\left(OH\right)_2}=0,1mol\\ m_{ddH_2SO_4}=\dfrac{0,1.98}{20\%}\cdot100\%=49g\\ V_{ddH_2SO_4}=\dfrac{49}{1,14}=42,98ml\)
PTHH: \(CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\)
Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)=n_{Ba\left(OH\right)_2}=n_{BaCO_3}\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\\m_{BaCO_3}=0,2\cdot197=39,4\left(g\right)\end{matrix}\right.\)
