By the use of stoichiometry, the mass of BaO in mixture is 0.06475 g.
<h3>What is the mass of BaO?</h3>
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.
Learn more about stoichiometry:brainly.com/question/9743981
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