Question

Answer 1

\(n_{SO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{NaOH}=0,1.2,2=0,22\left(mol\right)\)

\(T=\dfrac{0,22}{0,1}=2,2\rightarrow\) Tạo muối trung hoà và NaOH dư

PTHH: 2NaOH + SO₂ ---> Na₂SO₃ + H₂O

                            0,1           0,1

=> m_muối = 0,1.126 = 12,6 (g)

Answer 2

\(n_{SO_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(n_{NaOH}=0,1\cdot2,2=0,22mol\)

\(T=\dfrac{n_{OH^-}}{n_{SO_2}}=\dfrac{0,22}{0,1}=2,2\Rightarrow muốiNa_2SO_4\)

\(n_{Na_2SO_4}=0,1mol\Rightarrow m_{Na_2SO_4}=14,2g\)