\(n_{BaCO_3}=\dfrac{m}{M}=\dfrac{29,55}{197}=0,15\left(mol\right)\)
\(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\)
Theo đề ra : BaCO3 dư ( khỏi chứng minh ^_^)
PTHH
BaCO3 --to--> BaO + CO2
..0,12...............0,12......0,12
\(m_{BaO}=0,12\cdot153=22,95\left(g\right)\)
\(m_{BaCO_3}dư=\left(0,15-0,12\right)\cdot197=5,91\left(g\right)\)