Question

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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 115 torr, still at 30.0°C. Calculate the mass of BaO(s) in the initial mixture in grams. (Assume ideal gas behavior).

Answer 1

By the use of stoichiometry, the mass of BaO in mixture is 0.06475 g.

What is the mass of BaO?

From the question;

• V = 3.50 L
• T = 303 K
• P = 750 torr

Mass of BaO in the mixture = x grams

Mass of MgO in the mixture = (6.35 - x) grams

Number of  moles of CO2 initially present;

PV = nRT

= ((750/760) × 3.50) / 0.0821 × 303

n= 0.139

Number of moles of CO2 at the end;

n= PV /RT

= ((245/760) × 3.5) / 303 × 0.0821

= 0.045 mole

Amount of CO2 reacted;

= 0.139 - 0.045

= 0.044 mole

Now;

amount of reacted CO2 = ( amount of BaO + amount of MgO)

amount of BaO in mixture = x / 153 mole

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

Hence;

= 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 g

Hence, the mass of BaO in mixture is 0.06475 g.

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