BaO2 (s) + 2HCl (aq) → H2O2 (aq) + BaCl2 (aq) What mass of hydrogen peroxide should result when 1.50 g barium peroxide is treated with 25.0 mL hydrochloric acid solution containing 0.0272 g HCl per mL? 0.301 g H2O2 What mass of which reagent is left unreacted?

Question

Chemistry

Oscar Calhoun

7 March, 04:10

BaO2 (s) + 2HCl (aq) → H2O2 (aq) + BaCl2 (aq) What mass of hydrogen peroxide should result when 1.50 g barium peroxide is treated with 25.0 mL hydrochloric acid solution containing 0.0272 g HCl per mL? 0.301 g H2O2 What mass of which reagent is left unreacted?

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Answers (1)

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Abdiel Daniel · Answer 1

mass H2O2 = 0.31g H2O2

mass unreacted = 0.0745g BaO2

Explanation:

Mw BaO2 = 169.33 g / mol

Mw HCl = 36.46 g/mol

Mw H2O2 = 34.0147 g/mol

LR:

⇒ mol BaO2 = 1.50g BaO2 * mol / 169.33 g = 8.86E-3 mol BaO2 / 1 = 8.86E-3 mol BaO2

⇒ mol HCl = 25.0 mL * 0.0272g / mL * mol / 36.46g = 0.0186 mol HCl / 2 = 9.3E-3 mol HCl ... L. R

mol H2O2:

⇒ 0.0186mol HCl * (mol H2O2 / 2mol HCl) = 9.325E-3 mol H2O2

⇒ g H2O2 = 9.325E-3mol H2O2 * 34.0147g H2O2 / mol H2O2 = 0.31g H2O2

mass of wich reagent is left unreacted:

mol that react BaO2 = 0.0186mol HCl * (mol BaO2 / 2mol HCl) = 9.3E-3mol BaO2

mol that unreacted BaO2 = 9.3E-3 - 8.86E-3 = 4.4E-4 mol BaO2

g unreacted BaO2 = 4.4E-4mol BaO2 * (169.33g BaO2 / mol BaO2) = 0.0745g BaO2