Question

Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to the reaction BaO2(s)+H2SO4(aq)⟶BaSO4(s)+H2O2(aq) How many milliliters of 3.75 M H2SO4(aq) are needed to react completely with 53.5 g BaO2(s)?

Answer 1

Answer:

84.43 milliliters of 3.75 M Sulfuric acid are needed.

Explanation:

Moles of barium oxide = $\frac{53.5 g}{169 g/mol}=0.3166 mol$

$BaO_2(s)+H_2SO_4(aq)\rightarrow BaSO_4(s)+H_2O_2(aq)$

According to reaction, 1 mole of barium oxide reacts with 1 mole of sulfuric acid.

Then 0.3166 moles of barium oxide will react with:

$\frac{1}{1}\times 0.3166 mol=0.3166 mol$ of sulfuric acid.

$Molarity=\frac{\text{Moles of compound}}{V(L)}$

Where: V = Volume of the solution in Liters

Moles of sulfuric acid = 0.3166 mol

Volume of the sulfuric acid solution = V = ?

Molarity of sulfuric acid = 3.75 M

$3.75 m=\frac{0.3166 mol}{V}$

$V=\frac{0.3166 mol}{3.75 M}=0.08443 L$

0.08443 L = 84.43 mL (1 L = 1000 mL)

84.43 milliliters of 3.75 M Sulfuric acid are needed.

Answer 2

Answer:

VH2SO4 = 145.3 mL

Explanation:

Mw BaO2 = 169.33 g/mol

⇒ mol BaO2 = 53.5g * ( mol BaO2 / 169.33 g BaO2) = 0.545 mol BaO2

⇒according to the reaction:

mol BaO2 = mol H2SO4 = 0.545 mol

⇒ V H2SO4 = 0.545 mol H2SO4 * ( L H2SO4 / 3.75 mol H2SO4 )

⇒V H2SO4 = 0.1453 L (145.3 mL)