What volume of CO2 may be prepared from 5.00 lb CaCO3

Question

CaCO3——-> CaO+CO2

(STP)

Answer 1

Oleg,

Once we have the needed conversion factors and the molecular mass of calcium carbonate, the problems becomes a straight-forward dimensional analysis problem. The two conversion factors we need are as follows:

1.) 1 pound - 453.592 grams

2.) 1 mole = 22.4 Liters (at STP)

Mass of CaCO₃ = 1(40.08 g) + 1(12.01 g) + 3(16.00) = 40.08 g + 12.01 g + 48.00 g = 100.09 g

Next, we make sure the equation is balanced, which it is.

Now we set this us as a dimensional analysis problem:

5.00 lb CaCO₃ (453.592g/ 1 lb)(1 mol CaCO₃/ 100.09 g)(1 mol CO₂/1 mol CaCO₃) (22.4 L/1 mol) = 508 L

508 Liters will be produced.

Answer 2

Since we like to work in moles, we will convert 5.00 lbs of CaCO₃ to grams and then to moles.

molar mass CaCO₃ = 100.09 g/mole

1 lb = 454 grams

5.00 lbs x 454 g/lb x 1 mole/100. g = 22.7 moles of CaCO₃

moles CO₂ produced = 22.7 moles CaCO₃ x 1 mole CO₂/mole CaCO₃ = 22.7 moles CO₂ produced

At STP 1 mole = 22.4 liters

Volume CO₂ produced = 22.7 moles x 22.4 L/mole = 508 liters