\(n_{CO_2}=\frac{8,8}{44}=0,2\left(mol\right)\)
\(m_{NaOH}=200\times10\%=20\left(g\right)\)
\(\Rightarrow n_{NaOH}=\frac{20}{40}=0,5\left(mol\right)\)
CO2 + 2NaOH → Na2CO3 + H2O (1)
Theo pt1: \(n_{CO_2}=\frac{1}{2}n_{NaOH}\)
Theo bài: \(n_{CO_2}=\frac{2}{5}n_{NaOH}\)
Vì \(\frac{2}{5}< \frac{1}{2}\) ⇒ NaOH dư
Ta có: \(m_{dd}saupư=8,8+200=208,8\left(g\right)\)
Theo pT: \(n_{NaOH}pư=2n_{CO_2}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow n_{NaOH}dư=0,5-0,4=0,1\left(mol\right)\)
\(\Rightarrow m_{NaOH}dư=0,1\times40=4\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\frac{4}{208,8}\times100\%=1,92\%\)
Theo Pt: \(n_{Na_2CO_3}=n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Na_2CO_3}=0,2\times106=21,2\left(g\right)\)
\(\Rightarrow C\%_{Na_2CO_3}=\frac{21,2}{208,8}\times100\%=10,15\%\)
nCO2 = 0.2 mol
mNaOH = 20g
nNaOH = 0.5 mol
nNaOH/nCO2 = 0.5/0.2 = 2.5 => Tạo ra muối trung hòa
2NaOH + CO2 --> Na2CO3 + H2O
0.4______0.2_______0.2
mNaOH dư = ( 0.5 - 0.4 ) *40 = 4g
mNa2CO3 = 21.2 g
mdd sau phản ứng = 8.8 + 200 = 208.8 g
C%NaOH dư = 4/208.8*100% = 1.91%
C%Na2CO3 = 10.15%
