A 211 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