If 0.158 g of a white, unknown solid carbonate of a group 2a metal (m) is heated and the resulting co2 is transferred to a 285 ml sealed flask and allowed to cool to 25 degrees celsius, the pressure in the flask is 69.8 mmhg. what is the identity of the carbonate?

Using the ideal gas law equation we can find the number of moles of CO₂ formed 
PV = nRT
where 
P - pressure - 69.8 mmHg x 133 Pa/mmHg = 9 283 Pa
V - volume - 285 x 10⁻⁶ m³
n - number of moles 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 25 °C + 273 = 298 K
substituting these values in the equation 
9283 Pa x 285 x 10⁻⁶ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 298 K
n = 1.067 x 10⁻³ mol

decomposition of metal carbonate is as follows 
MCO₃ ---> MO + CO₂
stoichiometry of MCO₃ to CO₂ is 1:1
therefore number of moles of MCO₃ heated = number of CO₂ moles formed 
number of MCO₃ moles = 1.067 x 10⁻³ mol
molar mass = mass / number of moles 
molar mass = 0.158 g / 1.067 x 10⁻³ mol = 148 g/mol 
since carbonate molar mass is known -
 (molar mass of C x 1 C atom) + (molar mass of O x 3  O atoms)  
12 + 16x 3 = 12 + 48 = 60 
then mass of metal M - 148 - 60 = 88
group II metal with molar mass of 88 is Ra - Radium 

It will appear as o2(g) as a reactant