Tóm tắt:
$V_{SO_2}=112ml=0,112l$
$V_{Ca(OH)_2}=700ml=0,7l$
$C_{M\, Ca(OH)_2}=0,01M$
a/ $SO_2+Ca(OH)_2→CaSO_3+H_2O$
b/ $112ml=0,112l;700ml=0,7l$
$n_{Ca(OH)_2}=0,7.0,01=0,007(mol)$
$n_{SO_2}=\dfrac{0,112}{22,4}=0,005(mol)$
Theo phương trình: $\dfrac{n_{Ca(OH)_2}}{1}>\dfrac{n_{SO_2}}{1}(0,007>0,005mol)$
$→Ca(OH)_2$ dư $0,002mol$
Theo phương trình: $n_{SO_2}=n_{CaSO_3}=n_{H_2O}$
$→n_{CaSO_3}=n_{H_2O}=0,005mol$
$→\begin{cases}m_{CaSO_3}=0,6g\\m_{H_2O}=0,09g\end{cases}$
Vậy $m_{CaSO_3}=0,6g;\,m_{H_2O}=0,09g$