What amount of hydrogen peroxide should result when 1.21 g of barium peroxide is treated with 25.1 mL of hydrochloric acid solution containing 0.0289 g of HCl per mL?
1. Write the equation and balance it.
2a. Convert 1.21 g BaO2 to moles. mol = grams/molar mass.
2b. Do the same for the HCl. 0.0289 g/mL x 25.1 mL = grams, then mols = g/molar mass
3a. Using the coefficients in the balanced equation, convert mols BaO2 to moles of the product.
3b. Same for HCl
3c. You get two answers and only one can be right. The correct answer in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent.
4. Using the smaller value, convert that to g. g = mols x molar mass.
To find the amount of hydrogen peroxide produced when barium peroxide reacts with hydrochloric acid, we need to balance the equation and use stoichiometry.
First, let's write the balanced chemical equation:
BaO2 + 2HCl -> H2O2 + BaCl2
From the equation, we can see that one mole of barium peroxide (BaO2) reacts with two moles of hydrochloric acid (HCl) to produce one mole of hydrogen peroxide (H2O2) and one mole of barium chloride (BaCl2).
Now, let's calculate the moles of barium peroxide and hydrochloric acid:
Moles of BaO2 = mass of BaO2 / molar mass of BaO2
Moles of BaO2 = 1.21 g / (137.33 g/mol)
Moles of BaO2 = 0.00881 mol
Moles of HCl = volume of HCl solution (in L) x concentration of HCl
Moles of HCl = 25.1 mL x (0.0289 g/mL) / (36.46 g/mol)
Moles of HCl = 0.0198 mol
Since the stoichiometric ratio between BaO2 and H2O2 is 1:1, the moles of hydrogen peroxide produced will also be 0.00881 mol.
Finally, let's calculate the amount of hydrogen peroxide in grams:
Mass of H2O2 = moles of H2O2 x molar mass of H2O2
Mass of H2O2 = 0.00881 mol x (34.01 g/mol)
Mass of H2O2 = 0.299 g
Therefore, when 1.21 g of barium peroxide is treated with 25.1 mL of hydrochloric acid solution, the amount of hydrogen peroxide produced is approximately 0.299 grams.