A sample that contains only CaCOs and BaCOs weighs 5.40 g. Each compound reacts

Question

A sample that contains only CaCOs and BaCOs weighs 5.40 g. Each compound reacts in acid to liberate 1.39 L of C02(g), measured at 50& degree ;C and 0.904 atm. CaCOs(aq) rightarrow CaO(aq) + CO2(g) BaCO3(aq) rightarrow BaO(aq) + CO2(g) Assuming complete reactions, calculate the percentages by mass of CaCO3 and BaCO3 in the original mixture.

Explanation / Answer

let the weight of caco3 be x and that of BaCO3 be 5.40 - X. moles of CaCO3 = X / 100. moles of BaCO3 = 5.4 - X / 197. moles of Co2 produced should be equal to the sum of moles of CaCO3 and BaCO3. = X /100 + (5.4 - X) / 197. moles of co2 formed = 0.904*10^5*1.39 *10^-3/ (323*8.314) = 0.0467919 X /100 + (5.4 - X) / 197 = 0.047 => X = 2.936 weight of caco3 = 3.936 g weight of baco3 = 1.464 g. % weight of caco3 = (3.936 / 5.4 )*100 = 72.89% % weight of baco3 = (1.464 / 5.4 )*100 = 27.11%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.