Limiting Reactant Calculations

[Delilah Naranjo dis 3j]

The question is from the Post-Assessment Limiting Reactant Calculations #10.
How many moles of CO2(g) are produced when 500g of CaCO3(s) is used to neutralize an acid spill? the equation for the reaction at 1 atm and 25degrees C is:
CaCo3(S)+H2SO4(aq)----> CaSO4(s)+CO2+H2O(l)

What is 1atm and where does the 25 degrees C come into play?

[Rachel Wile 2D]

For this problem, I believe you can just ignore the physical conditions since they are not extreme. The only reason 1 atm and 25 degrees C are given in the problem is so the reader knows that the reaction is not put under extreme conditions, such as extremely high temperatures, which could cause H2SO4 to disassociate more.

As for how to do the problem, first you have to convert 500 g of CaCO3 to moles. The balanced chemical equation shows a 1:1 molar ratio between CaCo3 and CO2. So, one mole of CaCO3 makes one mole of CO2. Since the atomic mass of CaCO3 is about 100 g/mol, we know we have 5 moles of CaCO3, and therefore using our molar ratio we can produce 5 moles of CO2.