Quy đổi hh về \(\left\{{}\begin{matrix}Na\\Na_2O\\BaO\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\n_{Ba\left(OH\right)_2}=\dfrac{20,52}{171}=0,12\left(mol\right)\\n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\)
PTHH:
`2Na + 2H_2O -> 2NaOH + H_2`
`Na_2O + H_2O -> 2NaOH`
`BaO + H_2O -> Ba(OH)_2`
Theo PT: \(\left\{{}\begin{matrix}n_{Na}=2n_{H_2}=0,1\left(mol\right)\\n_{BaO}=n_{Ba\left(OH\right)_2}=0,12\left(mol\right)\end{matrix}\right.\)
`=>` \(n_{Na_2O}=\dfrac{21,9-0,12.153-0,1.23}{62}=0,02\left(mol\right)\)
Theo PT: \(\sum n_{NaOH}=2n_{Na_2O}+n_{Na}=0,09\left(mol\right)\)
PTHH: \(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3\downarrow+H_2O\)
bđ 0,12 0,3
pư 0,12-------->0,12
sau pư 0 0,18 0,12
\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
bđ 0,09 0,18
pư 0,09------>0,045
sau pư 0 0,135 0,045
\(Na_2CO_3+CO_2+H_2O\rightarrow2NaHCO_3\)
bđ 0,045 0,135
pư 0,045------>0,045
sau pư 0 0,09
\(BaCO_3+CO_2+H_2O\rightarrow Ba\left(HCO_3\right)_2\)
bđ 0,12 0,09
pư 0,09<----0,09
sau pư 0,03 0
`=>` \(m=m_{BaCO_3}=0,03.197=5,91\left(g\right)\)
