A. how many grams of co2 were produced?
We use the ideal gas equation,
PV = nRT
where P = pressure in atm (750/760)Â
      V = Volume  (75.0 mL)Â
      n = moles
      T = temperature in Kelvin (20+273.15=293.15)Â
Substitute the given values,
(750/760 atm) (0.075L)=(n) (0.0820 L-atm/mol K) (293.15 K)Â
n=0.00308 mol CO2
0.00308 (44.01 g / mol) = 0.136 g CO2Â
B. write the equations for the decomposition of both carbonates described above.
Thee decomposition reaction are as follows:
MgCO3 -> MgO + CO2Â
CaCO3 -> CaO + CO2Â
C. what percent of the limestone by mass was caco3?
Molar mass of CaCO3Â = 100.11g/molÂ
mole fraction  of Ca in CaCO3 = 40.1 / 100.11 = 0.401Â
1 / 0.401 (0.0448 g Ca)Â = 0.112 g CaCO3Â
%CaCO3 = 0.112 / 0.280 x 100 = 40%
100% - 40% CaCO3 = 60% MgCO3Â
D. how many grams of the magnesium- containing product were present?
0.6 * 0.2800g sample = 0.168g MgCO3Â
mole fraction MgO in MgCO3 =Â 40.3 / 84.31 = 0.477Â
0.477 mol fraction (0.168g MgCO3) = 0.080 g MgO in heated sample