To find the mass of hydrogen peroxide produced, we need to use stoichiometry, which involves converting the given quantities of barium peroxide and hydrochloric acid to moles and then using the balanced chemical equation to determine the mole ratio between barium peroxide and hydrogen peroxide.
1. Calculate the number of moles of barium peroxide:
Given mass of barium peroxide (BaO2) = 15.0 g
Molar mass of BaO2 = 169.34 g/mol (1 Ba + 2 O)
Number of moles of BaO2 = mass / molar mass = 15.0 g / 169.34 g/mol = 0.0886 mol
2. Calculate the number of moles of HCl:
Given volume of HCl solution = 25.0 mL
Given mass of HCl per mL = 0.0272 g/mL
Total mass of HCl = volume * mass per mL = 25.0 mL * 0.0272 g/mL = 0.68 g
Molar mass of HCl = 36.46 g/mol (1 H + 1 Cl)
Number of moles of HCl = mass / molar mass = 0.68 g / 36.46 g/mol = 0.0186 mol
3. Determine the mole ratio between BaO2 and H2O2:
From the balanced chemical equation: 1 mol BaO2 produces 1 mol H2O2
So, 0.0886 mol BaO2 will produce 0.0886 mol H2O2.
4. Convert the number of moles of H2O2 to mass:
Molar mass of H2O2 = 34.02 g/mol (2 H + 2 O)
Mass of H2O2 = number of moles * molar mass = 0.0886 mol * 34.02 g/mol = 3.02 g
Therefore, 3.02 grams of hydrogen peroxide should be produced.
To calculate the mass of each reagent left unreacted, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction, and the amount of the other reactant will be in excess.
5. Compare the mole ratio of the reactants:
The mole ratio of BaO2 to H2O2 is 1:1.
Therefore, the number of moles of BaO2 needed to produce 0.0886 mol H2O2 is also 0.0886 mol.
6. Compare the number of moles of BaO2 and HCl:
Number of moles for BaO2 = 0.0886 mol (from step 1)
Number of moles for HCl = 0.0186 mol (from step 2)
Since the number of moles of HCl is less than the number of moles of BaO2, HCl is the limiting reactant.
7. Determine the excess of BaO2:
The moles of BaO2 remaining = moles supplied - moles consumed
= 0.0886 mol - 0.0186 mol = 0.070 mol
Now, calculate the mass of unreacted BaO2:
Mass of unreacted BaO2 = moles remaining * molar mass
= 0.070 mol * 169.34 g/mol = 11.85 g
8. Determine the excess of HCl:
The moles of HCl remaining can be calculated using the mole ratio:
Mole ratio of HCl to BaO2 is 2:1
So, moles of HCl consumed = 2 * moles of BaO2 consumed = 2 * 0.0186 mol = 0.0372 mol
The moles of HCl remaining = moles supplied - moles consumed
= 0.0186 mol - 0.0372 mol = -0.0186 mol (negative value indicates excess amount)
Since there is an excess of HCl, the mass of unreacted HCl will be the difference between the total mass supplied and the mass consumed.
Mass of unreacted HCl = Total mass of HCl supplied - Mass of HCl consumed
= 0.68 g - (0.0186 mol * 36.46 g/mol) = 0.68 g - 0.677 g = 0.003 g
Therefore, the mass of each reagent left unreacted is 11.85 grams of BaO2 and 0.003 grams of HCl.