2NaOH + CO2 → Na2CO3 + H2O
\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow m_{CO_2}=0,1\times44=4,4\left(g\right)\)
\(m_{NaOH}=50\times24\%=12\left(g\right)\)
\(\Rightarrow n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{CO_2}\)
Theo bài: \(n_{NaOH}=3n_{CO_2}\)
Vì \(3>2\) ⇒ dd NaOH dư, CO2 hết
Dung dịch thu được sau phản ứng là: NaOH dư và Na2CO3
Theo PT: \(n_{NaOH}pư=2n_{CO_2}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow n_{NaOH}dư=0,3-0,2=0,1\left(mol\right)\)
\(\Rightarrow m_{NaOH}dư=0,1\times40=4\left(g\right)\)
Theo PT: \(n_{Na_2CO_3}=n_{CO_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Na_2CO_3}=0,1\times106=10,6\left(g\right)\)
\(\Sigma m_{dd}=4,4+50=54,4\left(g\right)\)
\(\Rightarrow C\%_{ddNa_2CO_3}=\dfrac{10,6}{54,4}\times100\%=19,49\%\)
\(C\%_{ddNaOH}dư=\dfrac{4}{54,4}\times100\%=7,35\%\)
