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