# How much CO2 is collected(in L) at STP on heating x g of Calcium Carbonete (CaCO3) ?

First of all pls tell me What STP is and also xplain the questions. Loads of thnx

Relevance

STP stands for 'standard temperature and pressure' and provides a unifying reporting system for thermochemical data. IUPAC currently defines this to be a temperature

T = 273.15 K and a pressure p = 100,000 Pa.

One thing to note is that at STP, one mole of an ideal gas occupies a volume called a standard molar volume.

Using the ideal gas equation:

pV = nRT ==> V = nRT/p

for one mole, n = 1

==> standard molar volume, V°m = (8.314)(273.15)/(100000)

V°m = 2.271 x 10ˉ² m³

= 22.71 L

Note that the OLD value of V°m was 22.4 L as the OLD standard pressure USED to be 1 atm = 101325 Pa. IUPAC no longer use 1 atm as standard pressure.

CaCO3(s) --> CaO(s) + CO2(g)

The balanced equation tells you that for every mole of CaCO3 that is completely decomposed, you get 1 mole of CO2 (as well as 1 mole of CaO).

The mass of one mole of CaCO3 is given by its molar mass and is 100.1 g molˉ¹. Therefore for x g of CaCO3 you will get (x/100.1) moles of CO2.

You know from above that one mole of (ideal) gas has a volume of 22.71 L at STP. Therefore for x g of CaCO3 completely decomposed you will collect (x/100.1)*(22.71) L of CO2 at STP.

• STP = standard temperature and pressure (273.15K and 100KPa)

You need to know the equation

CaCO3(s) --> CaO(s) + CO2(g)

So one mole of CaCO3 gives off one mole of CO2

At STP one mole of gas occupies exactly 22.4dm³ = 22.4L

100g CaCO3 = 1mol

So for every 100g of CaCO3 you will get 22.4L of CO2