CO2 + 2NaOH → Na2CO3 + H2O
\(n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\)
\(m_{NaOH}=200\times10\%=20\left(g\right)\)
\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
Theo PT: \(n_{CO_2}=\dfrac{1}{2}n_{NaOH}\)
Theo bài: \(n_{CO_2}=\dfrac{2}{5}n_{NaOH}\)
Vì \(\dfrac{2}{5}< \dfrac{1}{2}\) ⇒ NaOH dư
Dung dịch X gồm: NaOH dư và Na2CO3
\(m_{dd}saupư=8,8+200=208,8\left(g\right)\)
Theo pT: \(n_{NaOH}pư=2n_{CO_2}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow n_{NaOH}dư=0,5-0,4=0,1\left(mol\right)\)
\(\Rightarrow m_{NaOH}dư=0,1\times40=4\left(g\right)\)
\(\Rightarrow C\%_{NaOH}dư=\dfrac{4}{208,8}\times100\%=1,92\%\)
Theo PT: \(n_{Na_2CO_3}=n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Na_2CO_3}=0,2\times106=21,2\left(g\right)\)
\(\Rightarrow C\%_{Na_2CO_3}=\dfrac{21,2}{208,8}\times100\%=10,15\%\)
\(n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\)
\(m_{NaOH}=200\times10\%=20\left(g\right)\)
\(\Rightarrow n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
PTHH: CO2 + 2NaOH → Na2CO3 + H2O
Ban đầu: 0,2.........0,5..........................................(mol)
Phản ứng: 0,2.........0,4..........................................(mol)
Sau phản ứng: 0..........0,1........→......0,2...................(mol)
Dung dịch X gồm: NaOH dư và Na2CO3
\(m_{dd}saupư=8,8+200=208,8\left(g\right)\)
\(m_{NaOH}dư=0,1\times40=4\left(g\right)\)
\(\Rightarrow C\%_{NaOH}dư=\dfrac{4}{208,8}\times100\%=1,92\%\)
\(m_{Na_2CO_3}=0,2\times106=21,2\left(g\right)\)
\(\Rightarrow C\%_{Na_2CO_3}=\dfrac{21,2}{208,8}\times100\%=10,15\%\)
