Question

Answer 1

BT 4:

\(n_{SO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_{KOH}=0,15.1=0,15\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{KOH}}{n_{SO_2}}=\dfrac{0,15}{0,15}=1\) => Tạo muối KHSO₃

PTHH: KOH + SO₂ --> KHSO₃

             0,15------------->0,15

=> m_KHSO3 = 0,15.120 = 18 (g)

BT 5:

\(n_{SO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_{KOH}=0,2.1=0,2\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{KOH}}{n_{SO_2}}=\dfrac{0,2}{0,15}=1,33\) => Tạo muối K₂SO₃, KHSO₃

PTHH: 2KOH + SO₂ --> K₂SO₃ + H₂O

               0,2-->0,1------->0,1

            K₂SO₃ + SO₂ + H₂O --> 2KHSO₃

             0,05<---0,05-------------->0,1

=> Muối gồm \(\left\{{}\begin{matrix}K_2SO_3:0,05\left(mol\right)\\KHSO_3:0,1\left(mol\right)\end{matrix}\right.\)

=> m_muối = 0,05.158 + 0,1.120 = 19,9 (g)