Question

Dẫn 4,2l khí SO2 ( đktc) vào 275ml dd KOH 1M . Tính khối lượng muối thu được

 

Answer 1

\(n_{SO_2}=\dfrac{4,2}{22,4}=0,1875\left(mol\right)\)

n_KOH = 0,275 (mol)

PTHH: 2KOH + SO₂ --> K₂SO₃ + H₂O

_____0,275-->0,1375->0,1375

K₂SO₃ + SO₂ + H₂O --> 2KHSO₃

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

=> \(\left\{{}\begin{matrix}m_{KHSO_3}=0,1.120=12\left(g\right)\\m_{K_2SO_3}=\left(0,1375-0,05\right).158=13,825\left(g\right)\end{matrix}\right.\)

Answer 2

nSO2=4,2/22,4=0,1875(mol)nSO2=4,222,4=0,1875(mol)

n_KOH = 0,275 (mol)

PTHH: 2KOH + SO₂ --> K₂SO₃ + H₂O

_____0,275-->0,1375->0,1375

K₂SO₃ + SO₂ + H₂O --> 2KHSO₃

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

=> {mKHSO3=0,1.120=12(g)mK2SO3=(0,1375−0,05).158=13,825(g)