A sample of dolomitic limestone containing only caco3 and mgco3 was analyzed.
a.when a 0.2800 gram sample of this limestone was decomposed by heating, 75.0 milliliters of co2 at 750 mmhg and 20 degrees celcius were evovled. how many grams of co2 were produced.
b.write the equations for the decomposition of both carbonates described above.
c.it was also determined that the initial sample contained 0.0488 gram of calcium. what percent of the limestone by mass was caco3?
d.how many grams of the magnesium- containing product were present in the sample in (a) after it had been heated?

Respuesta :

A. how many grams of co2 were produced?
We use the ideal gas equation,
PV = nRT
where P = p
ressure 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