a) Ba(OH)2 + 2CO2 --> Ba(HCO3)2
b) \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\); \(n_{Ba\left(OH\right)_2}=0,075.1=0,075\left(mol\right)\)
PTHH: Ba(OH)2 + 2CO2-->Ba(HCO3)2
______0,075---->0,15----->0,075
=> \(C_{M\left(Ba\left(HCO_3\right)_2\right)}=\dfrac{0,075}{0,075}=1M\)