a) \(C\%_{ddNaOH}=\dfrac{12,4}{200}.100=6,2\%\)
\(n_{NaOH}=\dfrac{12,4}{40}=0,31\left(mol\right)\)
b) PTHH: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
Ta có: \(n_{CO_2}=\dfrac{0,31}{2}=0,155\left(mol\right)\\ \rightarrow V_{CO_2\left(đktc\right)}=0,155.22,4=3,472\left(l\right)\)
c) \(PTHH:NaOH+CO_2\rightarrow NaHCO_3\\ \rightarrow n_{CO_2}=n_{NaOH}=0,31\left(mol\right)\\ \rightarrow V_{CO_2\left(đktc\right)}=0,31.22,4=6,944\left(l\right)\)