At 650 k, the reaction mgco3(s)⇌mgo(s)+co2(g)mgco3(s)⇌mgo( s)+co2(g) has kp=0.026kp=0.026. a 10.0-l container at 650 k has 1.0 g of mgo(s) and co2co2 at p=0.0260 atmp=0.0260 atm. the container is then compressed to a volume of 0.100 l. find the mass of mgco3mgco3 that is formed.

0.4076 g

Explanation:

Kp is the equilibrium constant based on pressure and depends only on gas substances. For a generic reaction

aA + bB ⇄ cC + dD

, where pX is the pressure of X in equilibrium.

For the reaction Kp = pCO₂

pCO₂ = 0.026 atm

The system is in equilibrium at the beginning. The compression occurs at a constant temperature, so using Boyle's law

P1V1 = P2V2

0.026*10 = P2*0.1

P2 = 2.6 atm

The reaction will reach again the equilibrium, and pCO₂ = 0.026 atm, then the rest will form MgCO₃, which will be 2.6 - 0.026 = 2.574 atm.

By the ideal gas law:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm*L/mol*K), and T is the temperature.

2.574*0.1 = n*0.082*650

53.3n = 0.2574

n = 4.83x10⁻³ mol

The stoichiometry of the reaction is 1 mol of MgCO₃ for 1 mol of CO₂, so it will form 4.83x10⁻³ mol of MgCO₃  .

The molar mass is:

MgCO₃: 24 g/mol of Mg + 12 g/mol of C + 3*16 g/mol of O = 84 g/mol

The mass formed is the molar mass multiplied by the number of moles:

m = 84x4.83x10⁻³

m = 0.4076 g

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