Ta có:
\(\left\{{}\begin{matrix}n_{SO2}=\frac{6,72}{22,4}=0,3\left(mol\right)\\n_{KOH}=\frac{169.20}{100.56}=0,7\left(mol\right)\end{matrix}\right.\)
\(\frac{n_{KOH}}{n_{SO2}}==\frac{0,7}{0,3}=2,33\)
=> Tạo 1 muối K2SO3
\(2KOH+SO_3\rightarrow K_2SO_3+H_2O\)
__0,7____0,3_______0,3_______
\(\Rightarrow m_{K2SO3}=0,3.158=47,4\left(g\right)\)
\(\Rightarrow C\%_{dd\left(K2SO3\right)}=\frac{47,4}{196}.100\%=24,18\%\)
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