Respuesta :
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)
Answer:
84.43 milliliters of 3.75 M Sulfuric acid are needed.
Explanation:
Moles of barium oxide = [tex]\frac{53.5 g}{169 g/mol}=0.3166 mol[/tex]
[tex]BaO_2(s)+H_2SO_4(aq)\rightarrow BaSO_4(s)+H_2O_2(aq)[/tex]
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:
[tex]\frac{1}{1}\times 0.3166 mol=0.3166 mol[/tex] of sulfuric acid.
[tex]Molarity=\frac{\text{Moles of compound}}{V(L)}[/tex]
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
[tex]3.75 m=\frac{0.3166 mol}{V}[/tex]
[tex]V=\frac{0.3166 mol}{3.75 M}=0.08443 L[/tex]
0.08443 L = 84.43 mL (1 L = 1000 mL)
84.43 milliliters of 3.75 M Sulfuric acid are needed.
