1)
\(n_{SO2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\)
\(n_{NaOH}=2n_{SO2}=0,1.2=0,2\left(mol\right)\)
\(CM_{Naoh}=\frac{0,2}{0,1}=2M\)
2)
\(n_{CO2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{NaOH}=\frac{10}{23+17}=0,25\left(mol\right)\)
Vì nNaOH :nCO2=1 \(\Rightarrow\) Tạo ra muối axit
\(NaOH+CO_2\rightarrow NaHCO_3\)
\(n_{NaOH}=n_{CO2}=n_{NaHCO3}=0,25\left(mol\right)\)
\(\Rightarrow m_{NaHCO3}=0,25.\left(23+1+60\right)=21\left(g\right)\)