$m_{NaOH}=50.8\%=4g \\⇒n_{NaOH}=\dfrac{4}{40}=0,1mol \\a.PTHH : \\CO_2+2NaOH\to Na_2CO_3+H_2O \\Theo\ pt : \\n_{CO_2}=\dfrac{1}{2}.n_{NaOH}=\dfrac{1}{2}.0,1=0,05mol \\⇒V=V_{CO_2}=0,05.22,4=1,12l$
b.Chất rắn khan là Na2CO3.
Theo pt :
$n_{H_2O}=\dfrac{1}{2}.n_{NaOH}=\dfrac{1}{2}.0,1=0,05mol \\⇒m_{H_2O}=0,05.18=0,9g \\m_{dd\ spu}=5,04+0,9=5,94g \\⇒C\%_{Na_2CO_3}=\dfrac{5,04}{5,94}.100\%=84,85\%$