nSO2=2,24/22,4=0,1mol.
nCa(OH)2=0,15.0,5=0,075mol
Ta có nSO2/Ca(OH)2=1,33>1
->Xảy ra phản ứng
SO2+Ca(OH)2->CaSO3+H2O. (1)
SO2+CaSO3+H2O->Ca(HSO3)2 (2)
Gọi số moLSO2 (1)là x
Số mol SO2 (2) lầy
Ta có hệ pt
x+y=0,1
x=0,075
Giải đc y=0,025;x=0,075
mCaSO3=0,075.120=9g
MCa(HSO3)2=0,025.202= 5,05g
b,CMCaSO3=0,075/0,15=0,5
CMCa(HSO3)2=0,025/0,15=0,167M
nSO2=0,1(mol)
nCa(OH)2=0,075(mol)
tỉ số K: \(\dfrac{n_{SO_2}}{n_{Ca\left(OH\right)_2}}=\dfrac{0,1}{0,075}\approx1,33\)
tạo ra 2 muối
pt: Ca(OH)2 + SO2-> CaSO3 + H2O
vậy:0,075------->0,075--->0,075(mol)
pt Ca(OH)2 + 2SO2-> Ca(HCO3)2
vậy: 0,0125<----0,025--->0,0125(mol)
mmuói=0,075.90+0,0125.162=8,775(g)
b) CM CáO3=n/V=0,075/0,15=0,5(M)
CM Ca(HCO3)2=n/V=0,125/0,15=0,83(M
SO2 + Ca(OH)2 → CaSO3 + H2O (1)
\(n_{SO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{Ca\left(OH\right)_2}=0,15\times0,5=0,075\left(mol\right)\)
Theo PT1: \(n_{SO_2}=n_{Ca\left(OH\right)_2}\)
Theo bài: \(n_{SO_2}=\dfrac{4}{3}n_{Ca\left(OH\right)_2}\)
Vì \(\dfrac{4}{3}>1\) ⇒ SO2 dư ⇒ phản ứng tiếp:
SO2 + CaSO3 + H2O → Ca(HSO3)2 (2)
a) Theo PT1: \(n_{SO_2}pư=n_{Ca\left(OH\right)_2}=0,075\left(mol\right)\)
\(\Rightarrow n_{SO_2}dư=0,1-0,075=0,025\left(mol\right)\)
Ta có: \(n_{SO_2}dư\left(1\right)=n_{SO_2}\left(2\right)=0,025\left(mol\right)\)
Theo PT1: \(n_{CaSO_3}=n_{Ca\left(OH\right)_2}=0,075\left(mol\right)\)
Ta có: \(n_{CaSO_3}\left(1\right)=n_{CaSO_3}\left(2\right)=0,075\left(mol\right)\)
Theo PT2: \(n_{SO_3}=n_{CaSO_3}\)
Theo PT2: \(n_{SO_3}=\dfrac{1}{3}n_{CaSO_3}\)
Vì \(\dfrac{1}{3}< 1\) ⇒ CaSO3 dư
Theo PT2: \(n_{Ca\left(HCO_3\right)_2}=n_{SO_2}=0,025\left(mol\right)\)
\(\Rightarrow m_{Ca\left(HCO_3\right)_2}=0,025\times162=4,05\left(g\right)\)
Theo PT2: \(n_{CaSO_3}pư=n_{SO_3}=0,025\left(mol\right)\)
\(\Rightarrow n_{CaSO_3}dư=0,075-0,025=0,05\left(mol\right)\)
\(\Rightarrow m_{CaSO_3}dư=0,05\times120=6\left(g\right)\)
b) \(C_{M_{CaSO_3}}dư=\dfrac{0,05}{0,15}=0,33\left(M\right)\)
\(C_{M_{Ca\left(HSO_3\right)_2}}=\dfrac{0,025}{0,15}=0,167\left(M\right)\)
