Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to the reaction bao 2 ( s ) + h 2 so 4 ( aq ) ⟶ baso 4 ( s ) + h 2 o 2 ( aq ) bao2(s)+h2so4(aq)⟶baso4(s)+h2o2(aq) how many milliliters of 3.00 m h 2 so 4 ( aq ) 3.00 m h2so4(aq) are needed to react completely with 92.5 g bao 2 ( s ) ? 92.5 g bao2(s)?

There is 182 mL of 3.00 M H2SO4 needed

Explanation:

Step 1: Data given

Mass of BaO2 = 92.5 grams

Molarity of 3.00 M H2SO4

Step 2: The balanced equation

BaO2 ( s ) + H2SO4(aq ) ⟶ BaSO4(s) + H2O2(aq )

For 1 mole of BaO2  we need 1 mole of H2SO4 to produce 1 mole of BaSO4 and 1 mole H2SO2

Step 3: Calculate moles of  BaO2

Moles BaO2 = mass BaO2 / Molar mass BaO2

Moles BaO2 = 92.5 / 169.33 g/mol

Moles BaO2 = 0.546 moles

Step 4: Calculate moles of H2SO4

For 1 mole BaO2 we need 1 mole of H2SO4

For 0.546 moles of BaO2 we need 0.546 moles of H2SO4

Step 5: Calculate the volume of H2SO4 needed

Volume = moles / molarity

Volume = 0.546 mol/ 3.00 mol/L

Volume = 0.182 L = 182 mL

There is 182 mL of 3.00 M H2SO4 needed

no

explanation:

go away, people want answers, not to hook up

yes

ya te ayud espera