Question

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. After the reaction to form BaCO₃(s) and MgCO₃(s) was completed, the pressure of CO₂(g) remaining was 135 torr. Calculate the mass of BaO(s) in the mixture.

Answer 1

Initial number of moles of carbon dioxide is calculated by using Ideal gas equation.

PV = nRT

Given that Carbon dioxide is present in 3.5 L flask and 30°C and 750 torr.

Number of moles of Carbon dioxide = PV/RT

= 0.986 atm*3.5 L/(0.0821 lit atm/ K-mol) * 303 K

= 0.139 moles

Pressure of carbon dioxide after completion of the reaction is 135 torr.

Number of moles of carbon dioxide =

0.178 atm*3.5 L/(0.0821 lit atm/ K-mol) * 303 K

= 0.0249 moles

Number of moles of carbon dioxide reacted = 0.139 - 0.0249

= 0.114 moles

Molermans of nao : 153.33 let weight of Bao: a weight of Mg 0 : 6.35- Bao + co BaCO3 153.33great with I mole of cor x g reacts will ? of an No. of mole of = 183.33 Coy reprired 6.35-1 6.35-x MgO + Co2 Mg Cog 40g requers I mole of co 6.35-2 reguen ? of co2 No. of mole of coz repend = 0 Total no of moles of co, repede 153-33 40 + 6.357 = 0.114 19.25 + 70 1182.23-153-31 -olly 400 + 6.35 x 153.13 - 153-3] . 40x153.3] 973.65- 113.33x = 699.2 113.33x = 274.47 274.47 = 2.429. 113.13 a = SIGNATURE 2.42 6.35-2-42 = 3.99 so in the minha Mgo in the mint Amart of