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)?
To find out how many milliliters of 3.50 M H2SO4(aq) are needed to react completely with 12.1 g of BaO2(s), we need to determine the number of moles of BaO2(s) and then use stoichiometry to calculate the amount of H2SO4 needed.
1. Calculate the number of moles of BaO2(s):
The molar mass of BaO2 is:
Ba: 137.33 g/mol
O: 16.00 g/mol (There are two oxygen atoms in BaO2)
Molar mass of BaO2 = (137.33 g/mol) + (2 * 16.00 g/mol) = 169.33 g/mol
Now, calculate the number of moles of BaO2:
Number of moles = Mass / Molar mass
Number of moles = 12.1 g / 169.33 g/mol
2. Calculate the number of moles of H2SO4 needed:
According to the balanced chemical equation:
1 mole of BaO2 reacts with 1 mole of H2SO4
Therefore, the number of moles of H2SO4 needed is equal to the number of moles of BaO2.
3. Calculate the volume of 3.50 M H2SO4(aq) needed:
The molarity (M) is defined as moles of solute per liter of solution (mol/L).
Moles of solute = Molarity * Volume (in liters)
Volume (in liters) = Moles of solute / Molarity
To calculate the volume in milliliters, multiply the volume in liters by 1000.
Volume (in milliliters) = Volume (in liters) * 1000
Substituting the values:
Volume (in liters) = Number of moles / Molarity
Volume (in milliliters) = (Number of moles / Molarity) * 1000
Now, plug in the values and calculate the volume of 3.50 M H2SO4 needed:
Volume (in liters) = (12.1 g / 169.33 g/mol) / 3.50 mol/L
Volume (in milliliters) = [(12.1 g / 169.33 g/mol) / 3.50 mol/L] * 1000
Calculate the value to find the volume (in milliliters).
To find the number of milliliters of 3.50 M H2SO4(aq) needed to react completely with 12.1 g of BaO2(s), we first need to calculate the amount of BaO2 in moles.
Given:
Mass of BaO2 = 12.1 g
The molar mass of BaO2 is calculated by adding the atomic masses of its constituent elements:
Ba: 137.33 g/mol
O: 16.00 g/mol (each O atom)
2 O atoms present in BaO2, so the molar mass of BaO2 = 137.33 g/mol + (2 * 16.00 g/mol) = 169.33 g/mol
Now, we can calculate the amount of BaO2 in moles using the formula:
moles = mass / molar mass
moles = 12.1 g / 169.33 g/mol
moles = 0.0715 mol BaO2
According to the balanced chemical equation, the stoichiometric ratio between BaO2 and H2SO4 is 1:1. This means that for every 1 mole of BaO2, we need 1 mole of H2SO4.
Since the molar concentration of H2SO4 is given as 3.50 M, we know that in 1 liter of solution, there are 3.50 moles of H2SO4.
To find the volume of H2SO4 solution needed, we can use the equation:
volume (in liters) = moles / molarity
volume = 0.0715 mol / 3.50 mol/L
volume = 0.0204 L
Finally, we convert the volume from liters to milliliters:
volume (in mL) = volume (in L) * 1000
volume = 0.0204 L * 1000
volume = 20.4 mL
Therefore, 20.4 milliliters of 3.50 M H2SO4(aq) are needed to react completely with 12.1 g of BaO2(s).
Convert 12.1 g BaO2 to mols. mol = g/molar mass
Use the coefficients in the balanced equation to convert mols BaO2 to mols H2SO4.
Then M H2SO4 = mols H2SO4/L H2SO4. You know M and mols solve for L and convert to mL.