Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to
BaO2+H2SO4 -> BaSO4 +H2O2
How many milliliters of 3.50 M H2SO4(aq) are needed to react completely with 12.1 g of BaO2(s)?
Duplicate post.
To determine the number of milliliters of 3.50 M H2SO4(aq) needed to react completely with 12.1 g of BaO2(s), we need to follow a few steps:
Step 1: Determine the number of moles of BaO2(s).
First, calculate the molar mass of BaO2:
1 Ba atom = 1 * 137.33 g/mol = 137.33 g/mol (from the periodic table)
2 O atoms = 2 * 16.00 g/mol = 32.00 g/mol (from the periodic table)
Total molar mass of BaO2 = 137.33 g/mol + 32.00 g/mol = 169.33 g/mol
Next, calculate the number of moles of BaO2:
moles = mass / molar mass = 12.1 g / 169.33 g/mol = 0.0715 mol
Step 2: Determine the stoichiometric ratio between BaO2 and H2SO4.
From the balanced equation, we see that the stoichiometric ratio between BaO2 and H2SO4 is 1:1. This means that 1 mol of BaO2 reacts with 1 mol of H2SO4.
Step 3: Calculate the number of moles of H2SO4 needed.
Since the stoichiometric ratio is 1:1, the number of moles of H2SO4 needed is also 0.0715 mol.
Step 4: Calculate the volume of 3.50 M H2SO4.
The concentration of 3.50 M means that for every 1 liter of solution, there are 3.50 moles of H2SO4.
To calculate the volume, we can use the formula:
volume (in liters) = moles / concentration
volume = 0.0715 mol / 3.50 mol/L = 0.0204 L
Finally, convert the volume to milliliters:
Volume (in milliliters) = volume (in liters) * 1000
Volume = 0.0204 L * 1000 = 20.4 mL
Therefore, approximately 20.4 milliliters of 3.50 M H2SO4(aq) are needed to react completely with 12.1 g of BaO2(s).