\(1/n_{SO_2}=\dfrac{7,437}{24,79}=0,3mol\\ n_{KOH}=0,25.2=0,5mol\\ T=\dfrac{0,5}{0,2}=2,5\)
→Tạo muối trung hòa, KOH dư\(\left(K_2SO_3\right)\)
\(2KOH+SO_2\rightarrow K_2CO_3+H_2O\)
0,4 0,2 0,2
\(m_{dd}=250.1,2+0,3.64=319,2g\\ C_{\%K_2SO_3}=\dfrac{0,2.138}{319,2}\cdot100\%=8,65\%\\ C_{\%KOH}=\dfrac{\left(0,5-0,4\right).56}{319,2}\cdot100\%=1,75\%\)
\(2,BaCl_2+K_2SO_3\rightarrow BaSO_3+2KCl\\ n_{\downarrow}=n_{BaSO_3}=n_{K_2SO_3}=0,2mol\\ m_{\downarrow}=m_{BaSO_3}=0,2.217=43,4g\)
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