At 650 K, the reaction MgCO3(s)?MgO(s)+CO2(g) has Kp=0.026. A 12.8L container at 650 K has 1.0g...

At 650 K, the reaction MgCO3(s)?MgO(s)+CO2(g) has Kp=0.026. A 12.8L container at 650 K has 1.0g of MgO(s) and CO2 at P = 0.0260 atm. The container is then compressed to a volume of 0.900L .

Find the mass of MgCO3 that is formed.

Expert Solution

Approximate .49 gram

K_{sp =} P_CO2

Case 1

P_co₂= nRT/V

V= 12.8 R = .082

P_{CO2 =} .00624

Case 2

P_{co2 =}= .00044 where V = 0.9 after appling above mentione formula

MgCO₃ ---------------> MgO + CO₂

.041 Mole .00624 Mole Initialy

(Limiting Reactant for back reaction)

.00582 .00044 Mole Finally

wherer .0058 = .00624 - .00044

so final produced weight of MgCO₃ is.0058 x (24 + 12 + 48) = 0.49 gram

queen_honey_blossom answered 1 month ago

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