Answer:
            Mass  =  47.04 g 
            Volume  =  23.94 L 
Solution:
The equation for given reaction is as follow,
                  BaCO₃  +  2 HNO₃     →     Ba(NO₃)₂  +  CO₂  +  H₂O
According to this equation,
            197.34 g (1 mole) BaCO₃ produces  =  44 g (1 mole) of CO₂
So,
                       211 g of BaCO₃ will produce  =  X g of CO₂
Solving for X,
                     X  =  (211 g × 44 g) ÷ 197.34 g
                     X  =  47.04 g of CO₂
As we know,
                  44 g (1 mole) CO₂ at STP occupies  =  22.4 L volume
So,
                                47.04 g of CO₂ will occupy  =  X L of Volume
Solving for X,
                     X  =  (47.04 g × 22.4 L) ÷ 44 g
                     X  =  23.94 L Volume