SO2 + 2NaOH → Na2SO3 + H2O
\(n_{SO_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{NaOH}=0,5\times2,5=1,25\left(mol\right)\)
a) Theo pT: \(n_{SO_2}=\frac{1}{2}n_{NaOH}\)
Theo bài: \(n_{SO_2}=\frac{6}{25}n_{NaOH}\)
Vì \(\frac{6}{25}< \frac{1}{2}\) ⇒ NaOH dư
Theo pT: \(n_{NaOH}pư=2n_{SO_2}=2\times0,3=0,6\left(mol\right)\)
\(\Rightarrow n_{NaOH}dư=1,25-0,6=0,65\left(mol\right)\)
\(\Rightarrow m_{NaOH}dư=0,65\times40=26\left(g\right)\)
b) DD sau pứ: NaOH dư và Na2SO3
\(C_{M_{NaOH}}dư=\frac{0,65}{0,5}=1,3\left(M\right)\)
Theo pT: \(n_{Na_2SO_3}=n_{SO_2}=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{Na_2SO_3}}=\frac{0,3}{0,5}=0,6\left(M\right)\)
nSO2= 6.72/22.4 = 0.3 mol
nNaOH = 0.5*2.5 = 1.25 mol
a) 2NaOH + SO2 --> Na2SO3 + H2O
Bđ: 1.25_____0.3
Pư :0.6______0.3______0.3
Kt: 0.65_____0.3______0.3
mNaOH dư = 0.65*40=26g
CM NaOH(dư) = 0.65/0.5=1.3M
CM Na2SO3 = 0.3/0.5 = 0.6M
