A211 g sample of barium carbonate, baco3, reacts with a solution of nitric acid to give barium nitrate, carbon dioxide, and water. if the acid is present in excess, what mass and volume of dry carbon dioxide gas at stp will be produced?

            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

There is one s orbital, and there are three p orbitals, five d orbitals, and seven f orbitals. thus, the answer is b. do not confuse the number of orbitals in a subshell with the number of electrons the subshell can hold. each orbital can hold two electrons, so the capacity of an nf subshell is 14 electrons.