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 175 torr, still at 30.0°C. Calculate the mass of BaO(s) in the initial mixture. (Assume ideal gas behavior).
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Answer:
please explain your question then i will answer it
Mass of BaO in the initial mixture is 3.50g
Volume = 3.50 L (Given)
Temp = 303 K (Given)
Pressure = 750 torr (Given)
Let the mass of BaO in the mixture be = x grams
Let the mass of MgO in the mixture be = (6.35 - x) grams
Initially, the pressure is at CO2
According to the ideal gas equation -
PV = nRT
= ((750/760) × 3.50) / 0.0821 × 303
= 0.139
Thus, CO2 is 0.139 moles
Final mole of CO2
n= PV /RT
= ((245/760) × 3.5) / 303 × 0.0821
= 0.045 mole
Reacted mole = 0.139 - 0.045
= 0.044 mole
Moles of reacted CO2 = ( moles of BaO + moles of MgO)
Moles of BaO in mixture = x / 153 mole
Moles of MgO in mixture = 6.35 - x mole / 40
On equating both, -
= 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