In the reaction: <span>caco3(s) → cao(s) + co2(g), it is evident that
1 mol (which is 100 g) of CaCO3 yields 1 mol (which is 44 g) of CO2
Now, the molarity of CaCO3 present in the reaction system is
</span>=

=

= 0.45 mol
Thus, 0.45 mol of CaCO3 leads to the formation of 0.45 mol of CO2.
According to the ideal gas equation, we have PV = nRT
V =

.
Considering P = 645 torr = 0.8487 atm (because 1 atm = 760 torr)
In that case, V =

= 34.8 l
1 percent of the solar radiation
The percentage of calcium carbonate that reacted is 2.5%. The reaction in question allows us to determine the equilibrium Kp: Kp = the partial pressure of carbon dioxide, since the other components are solids. We'll apply the ICE table to the provided equilibrium. At the start, we have 0.2 for calcium carbonate with no initial moles of other substances. As the reaction progresses, we set the changes to be -x for calcium carbonate, +x for carbon dioxide, and +x for the other product, leading us to an equilibrium of 0.2-x for calcium carbonate while both other products are at x. Using Kp = Kc(RT)ⁿ, where n represents the mole difference of gaseous products and reactants, we find n to equal 1 for this reaction. With R as the gas constant (8.314 J/mol K) and the temperature at 800 °C (1073 K), we substitute the values accordingly. Upon calculation, we find x = 0.005, which indicates the amount of calcium carbonate that dissociated or reacted, leading us to the reacted percentage.