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 3.00 M H2SO4(aq) are needed to react completly with 85.1g of BaO2(s)?
Here is an example of a stoichiometry problem. Just follow the steps. When you get to the end and have moles, remember
M = moles/L of soln.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the number of milliliters of 3.00 M H2SO4(aq) needed to react completely with 85.1g of BaO2(s), we can use stoichiometry and the concept of mole-to-mole ratios.
First, let's calculate the number of moles of BaO2(s) using its molar mass:
Molar mass of BaO2 = molar mass of Ba + 2 * molar mass of O = 137.33 g/mol + 32.00 g/mol = 169.33 g/mol
Moles of BaO2 = mass of BaO2 / molar mass of BaO2 = 85.1 g / 169.33 g/mol = 0.502 mol
Next, we need to determine the stoichiometric ratio between BaO2(s) and H2SO4(aq) using the balanced equation:
1 mol of BaO2 reacts with 1 mol of H2SO4.
Now, we can use the stoichiometry to find the number of moles of H2SO4 needed:
Moles of H2SO4 = 0.502 mol of BaO2 * (1 mol of H2SO4 / 1 mol of BaO2) = 0.502 mol
Finally, we can calculate the volume of the 3.00 M H2SO4(aq) solution needed, using the molar concentration formula:
Molar concentration (M) = moles / volume (in L)
We have the molar concentration (3.00 M) and the moles of H2SO4 (0.502 mol). Let's compute the volume in L first and then convert it to milliliters (mL):
Volume (L) = moles / molar concentration
Volume (L) = 0.502 mol / 3.00 mol/L = 0.167 L
To convert to milliliters:
Volume (mL) = Volume (L) × 1000 mL/L
Volume (mL) = 0.167 L × 1000 mL/L = 167 mL
Therefore, 167 milliliters of 3.00 M H2SO4(aq) are needed to react completely with 85.1g of BaO2(s).