\(n_{SO_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(m_{NaOH}=\frac{20.150}{100}=30\left(g\right)\)
=>\(n_{NaOH}=\frac{30}{40}=0,75\left(mol\right)\)
Ta thấy: \(\frac{n_{NaOH}}{n_{SO_2}}=\frac{0,75}{0,15}=5>2\)
=>SO2 tác dụng với dd NaOH tạo ra muối Na2SO3
\(SO_2+2NaOH\rightarrow Na_2SO_3+H_2O\)
0,15___0,3_______0,15
So sánh tỉ lệ:
\(\frac{n_{SO_2}}{1}< \frac{n_{NaOH}}{2}\Leftrightarrow\frac{0,15}{1}< \frac{0,75}{2}\)
=>NaOH dư
=>\(n_{NaOH\left(dư\right)}\)=0,75-0,3=0,45(mol)
=>\(m_{NaOH\left(dư\right)}=0,45.40=18\left(g\right)\)
mdd=\(m_{SO_2}+m_{ddNaOH}=0,15.64+150=159,6\left(g\right)\)
\(m_{Na_2SO_3}=0,15.126=18,9\left(g\right)\)
\(C\%_{Na_2SO_3}=\frac{18,9.100}{159,6}\approx11,84\%\)
\(C\%_{NaOH\left(dư\right)}=\frac{18.100}{159.6}=11,28\%\)
