How many liters of co(g) at stp are produced when 68.0 g of caco3(s) is heated according to the following equation? caco3(s) cao(s) + co2(g) : 15.2 l 0.679 l 68.0 l 30.4 l

15.23 L

Solution:

The balanced chemical equation is as,

CaCO₃ → CaO + CO₂

As at STP one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,

100 g ( 1 mol) CaCO₃ produces = 22.4 L (1 mol) of CO₂

So,

68 g CaCO₃ will produce = X L of CO₂

Solving for X,

X = (68 g × 22.4 L) ÷ 100 g

X = 15.23 L of CO₂