A chemist weighed out 5.14 g of a mixture containing unknown amounts of bao (s) and cao (s) and placed the sample in a 1.50-l flask containing co2 (g) at 30.0 oc and 750. torr. after the reaction to form baco3 (s) and caco3 (s) was completed, the pressure of co2 (g) remaining was 230. torr. calculate the mass percentages of cao (s) and bao (s) in the mixture. 86.6 % bao; 13.4 % cao

Let the mass of CaO = x grams

So mass of BaO = 5.14 - x grams

moles of CaO = mass / molar mass = x / 56

Moles of BaO = mass / molar mass = 5.14-x / 153

initial moles of CO2 = PV / RT = 750 X 1.50 / 760 X 0.0821 X 303 = 0.06

final mole sof CO2 = PV / RT = 230 X 1.50 / 760 X 0.0821 X 303 = 0.018

So moles of BaCO3 and CaCO3 formed = 0.06 - 0.018 = 0.042 moles

x / 56 + (5.14-x) / 153 = 0.042

on solving

x = 0.68

So mass of CaO = 0.68 g

So percentage of CaO = 0.68 X 100 / 5.14 = 13.4 %

Percentage of BaO = 86.6%