\(n_{C_4H_{10}}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\)
\(n_{Ba\left(OH\right)_2}=0,4.0,2=0,08\left(mol\right)\)
\(C_4H_{10}+\dfrac{13}{2}O_2\rightarrow4CO_2+5H_2O\left(1\right)\)
\(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3\downarrow+H_2O\left(2\right)\)
\(BaCO_3+H_2O+CO_2\rightarrow Ba\left(HCO_3\right)_2\left(3\right)\)
\(\left(1\right)\Rightarrow n_{CO_2}=0,025.4=0,1\left(mol\right)\)
\(\left(2\right)\Rightarrow n_{BaCO_3}=n_{Ba\left(OH\right)_2}=n_{CO_2}=0,08\left(mol\right)\Rightarrow n_{CO_2}\left(dư\right)=0,1-0,08=0,02\left(mol\right)\)
\(\left(3\right)\Rightarrow n_{BaCO_3}=0,02\left(mol\right)\)
\(n_{BaCO_3}\left(dư\right)=0,08-0,02=0,06\left(mol\right)\)
\(m_{BaCO_3}=0,06.197=11,82\left(g\right)\)