Hydrogen peroxide is used as a cleaning agent in the treatment of cuts and abrasions for several reasons. it is an oxidizing agent that can directly kill many microorganisms; it decomposes upon contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams upon contact with blood, which provides a cleansing action. in the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide. bao2(s) + 2 hcl(aq) h2o2(aq) + bacl2(aq) what amount of hydrogen peroxide should result when 1.49 g of barium peroxide is treated with 25.1 ml of hydrochloric acid solution containing 0.0279 g of hcl per ml? webassign will check your answer for the correct number of significant figures. 0.299 correct: your answer is correct. g what mass of which reagent is left unreacted? webassign will check your answer for the correct number of significant figures. 0.0367 incorrect: your answer is incorrect. g hcl bao2 correct: your answer is correct.

Respuesta :

Given:

BaO2(s) + 2 HCl(aq) => H2O2(aq) + BaCl2(aq) 

Mass of barium peroxide = 1.49 g Volume of HCl solution = 25.1 mLConcentration of HCl solution = 0.0279 g HCl per mL

Solution: To determine the amount of hydrogen peroxide that would be produced, we use need to determine which is the limiting reactant and use the initial amount of this reactant for the calculations. 

0.0279 g HCl / mL ( 25.1 mL ) ( 1 mol HCl / 36.46 g HCl) ( 1 mol BaO2 / 2 mol HCl ) = 0.0096 mol BaO2 needed
1.49 g BaO2 ( 1 mol / 169.3 g BaO2 ) ( 2 mol HCl / 1 mol BaO2 ) = 0.0176 mol HCl needed

Therefore, the limiting reactant would be barium peroxide since it is consumed completely first in the reaction.

1.49 g BaO2 ( 1 mol BaO2 / 169.3 g BaO2 ) ( 1 mol H2O2 / 1 mol BaO2 ) ( 34.02 g H2O2 / 1 mol H2O2 ) = 0.299 g H2O2