a) CO2 + Ba(OH)2 -------> BaCO3 + H2O
b) \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{Ba\left(OH\right)_2}=n_{CO_2}=0,1\left(mol\right)\)
=> \(CM_{Ba\left(OH\right)_2}=\dfrac{0,1}{0,1}=1M\)
c) \(n_{BaCO_3}=n_{CO_2}=0,1\left(mol\right)\)
=> \(m_{BaCO_3}=0,1.197=19,7\left(g\right)\)
10.\(n_{SO_2}=\dfrac{0,056}{22,4}=2,5.10^{-3}\left(mol\right)\)
\(n_{Ca\left(OH\right)_2}=0,35.0,01=3,5.10^{-3}\left(mol\right)\)
\(\dfrac{n_{OH^-}}{n_{SO_2}}=\dfrac{3,5.2.10^{-3}}{2,5.10^{-3}}=2,8\)
=>Chỉ tạo muối trung hòa, Ca(OH)2 dư
SO2 + Ca(OH)2 -------> CaSO3 + H2O
0,0025--->0,0025--------->0,0025
\(m_{CaSO_3}=0,0025.120=0,3\left(g\right)\)
\(m_{Ca\left(OH\right)_2\left(dư\right)}=\left(0,0035-0,0025\right).94=0,094\left(g\right)\)
Bài 9:
nCO2=0,1(mol)
a) PTHH: CO2 + Ba(OH)2 -> BaCO3 + H2O
b) nBa(OH)2=nBaCO3=nCO2=0,1(mol)
CMddBa(OH)2=0,1/0,1=1(M)
c) mBaCO3=0,1.197=19,7(g)
