Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to

BaO2(s)+H2SO4(aq)=BaSO4(s)+H2O2(aq)
How many milliliters of 5.00 M H2SO4(aq) are needed to react completely with 30.5 g of BaO2(s)?

found the answer to be 36.03 ml

36.03 mL is right but if your prof is picky about the number of significant figures the last three will count against you. You are allowed only 3 places or 36.0 mL.

To determine the number of milliliters of 5.00 M H2SO4(aq) needed to react completely with 30.5 g of BaO2(s), we need to use stoichiometry and the concept of molarity.

Here are the steps to solve the problem:

Step 1: Calculate the molar mass of BaO2(s).
The molar mass of BaO2(s) can be calculated by adding the atomic masses of barium (Ba) and oxygen (O).
Ba: 1 x 137.33 g/mol = 137.33 g/mol
O: 2 x 16.00 g/mol = 32.00 g/mol
Total molar mass = 137.33 g/mol + 32.00 g/mol = 169.33 g/mol

Step 2: Convert the mass of BaO2(s) to moles.
To convert grams to moles, divide the mass by the molar mass.
30.5 g BaO2(s) / 169.33 g/mol = 0.1802 mol BaO2(s)

Step 3: Determine the stoichiometric ratio between BaO2(s) and H2SO4(aq).
According to the balanced chemical equation:
1 mol BaO2(s) reacts with 2 mol H2SO4(aq)

Step 4: Calculate the number of moles of H2SO4(aq) needed.
Multiply the moles of BaO2(s) by the stoichiometric ratio.
0.1802 mol BaO2(s) x (2 mol H2SO4(aq) / 1 mol BaO2(s)) = 0.3604 mol H2SO4(aq)

Step 5: Calculate the volume of 5.00 M H2SO4(aq) required.
Use the formula for molarity:
Molarity (M) = moles / volume (L)
Rearranging the formula, we have:
Volume (L) = moles / Molarity (M)

Volume (L) = 0.3604 mol H2SO4(aq) / 5.00 mol/L = 0.0721 L

Step 6: Convert the volume from liters to milliliters.
Since there are 1000 mL in 1 L, multiply the volume in liters by 1000.
0.0721 L x 1000 mL/L = 72.1 mL

Therefore, 72.1 milliliters of 5.00 M H2SO4(aq) are needed to react completely with 30.5 g of BaO2(s).