Question

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

Answer 1

The answer is: A/ 15.2 L

Hope this helps!

Answer 2

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