SO2 + Ca(OH)2 -> CaSO3 + H2O
a................................a (mol)
với a = V/22,4 (mol)
Theo bài ra :
\(\Delta m=m_{CáSO3}-m_{SO2}=120a-64a=56a\) = 5,6 (g)
=> a = 0,1(mol)
=> V =2,24(l)
Gọi số mol SO2 là x(mol)
SO2+ Ca(OH)2---> CaSO3+ H2O
x...................................x.(mol)
Có: mCaSO3-mSO2=56x=5,6(g)
=> x=0,1(mol)
=> V=0,1.22,4=2,24(l)
Đặt :
nSO2 = x mol
Ca(OH)2 + SO2 --> CaSO3 + H2O
___________x_______x
mdd giảm = mKt - mSO2 = 5.6
<=> 120x - 64x = 5.6
<=> x = 0.1
VSO2 = 2.24 l
\(n_{SO_2}=x\)
\(TH_1:SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3+H_2O\)
(mol) x x x x
\(m_{SO_2}+m_{ddCa\left(OH\right)_2}-m_{ddCaSO_3}=5,6\\ \Leftrightarrow64x+5,6=120x\\ \Leftrightarrow x=\frac{5,6}{120-64}=0,1\left(mol\right)\\ \rightarrow V_{SO_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)
