$m_{NaOH}=50.8\%=4g \\⇒n_{NaOH}=\dfrac{4}{40}=0,1mol \\a.PTHH : \\CO_2+2NaOH\to Na_2CO_3+H_2O \\Theo\ pt : \\n_{CO_2}=\dfrac{1}{2}.n_{NaOH}=\dfrac{1}{2}.0,1=0,05mol \\⇒V=V_{CO_2}=0,05.22,4=1,12l$

b.Chất rắn khan là Na2CO3.

Theo pt :

$n_{H_2O}=\dfrac{1}{2}.n_{NaOH}=\dfrac{1}{2}.0,1=0,05mol \\⇒m_{H_2O}=0,05.18=0,9g \\m_{dd\ spu}=5,04+0,9=5,94g \\⇒C\%_{Na_2CO_3}=\dfrac{5,04}{5,94}.100\%=84,85\%$

Đáp án:

Giải thích các bước giải:

 nnaoh=50/23+16+1 x 8%=0.1 mol   

a/ ptpu:2NaOH+CO₂→H₂O+Na₂CO₃

=> nco2=1/2nnaoh=0.05mol

=>vco2=0.05.22.4=1.12l

b/ chất rắn khan na2co3

 nh2o=nco2=0.05

=>mh2o=0.05x18=0.9g

=>mdd=0.9+5.04=5.94g

C%Na₂CO_{3=5.04x100/5.94=85%}