\(n_{SO_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(n_{Ba\left(OH\right)_2}=\frac{342.15}{100.171}=0,3\left(mol\right)\rightarrow n_{OH^-}=0,6\left(mol\right)\)
Tỉ lệ: \(\frac{n_{OH}}{n_{SO_2}}=\frac{0,6}{0,2}=3>2\rightarrow\) Muối trung hòa
\(PTHH:Ba\left(OH\right)_2+SO_2\rightarrow BaSO_3+H_2O\)
(mol)______0,2_______0,2_______0,2____0,2_
\(m_{\downarrow}=0,2.217=43,4\left(g\right)\)
\(ddX:Ba\left(OH\right)_2\cdot du\cdot0,3-0,2=0,1\left(mol\right)\)
\(C\%_{Ba\left(OH\right)_2\cdot du}=\frac{0,1.171}{0,2.64+342-43,4}.100\%=5,5\left(\%\right)\)
a) \(n_{SO_2}=0,2\left(mol\right);n_{Ba\left(OH\right)_2}=0,3\left(mol\right)\)
Ba(OH)2 + SO2 ----> BaSO3 + H2O
0,2________0,2________0,2
\(\Rightarrow m_{BaSO_3}=43,4\left(g\right)\)
b) C%(Ba(OH)2) = 5,5(g)
