You placed 6.35 g of a mixture containing unknown amounts of BaO(s) and MgO(s) in a 3.50-L flask containing CO₂(g) at 30.0°C and 750. torr. The solid BaO and MgO in the flask completely reacted to form BaCO₃(s) and MgCO₃(s), respectively. After the reactions to form BaCO₃(s) and MgCO₃(s) were completed, the pressure of CO₂(g) remaining was 215 torr, still at 30.0°C. Calculate the mass of BaO(s) in the initial mixture. (Assume ideal gas behavior).

Answer by Guest

Mass of BaO in  initial mixture = 3.50g

Explanation:

Let mass of BaO in mixture be x g

mass of MgO in mixture be (6.35 - x) g

Initially CO_2

Volume = 3.50 L

Temp = 303 K

Pressure = 750 torr = 750 / 760 atm

Applying ideal gas equation

PV = nRT

n = PV / RT

(n)_CO_2 = ((750/760)* 3.50) / 0.0821 * 303

(n)_CO_2 = 0.139 mole

Finally; mole of CO_2

n= PV /RT

((245/760) *3.5) / 303* 0.0821

(n)_CO_2 = 0.045 mole

Mole of CO_2 reacted = 0.139 - 0.045

=0.044 mole

BaO + CO_2  BaCO_3

Mgo + CO_2  MgCO_3

moles of CO_2 reacted = ( moles of BaO + moles of MgO)

moles of BaO in mixture = x / 153 mole

moles of MgO in mixture = 6.35 - x mole / 40

Equating,

x/ 153 +6.35/40 = 0.094

= x/153 + 6.35 / 40 - x/40 =0.094

= x (1/40 - 1153) = (6.35/40 - 0.094)

= x * 10.018464

= 0.06475

mass of BaO in mixture = 3.50g

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