Small quantites of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide.
BaO2 (s) + 2HCl (aq) -> H2O2 (aq) + BaCl2 (aq)
What mass of hydrogen peroxide should result when 15.0g of barium peroxide is treated with 25.0 mL of hydrochloric acid solution containing 0.0272g of HCl per mL? What mass of each reagent is left unreacted?
Thanks!
So you have 15.0g BaO2 and 0.0272 g/mL x 25.0 mL = grams HCl. This is a limiting reagent and you must first convert grams to mols. Here is a worked example of a limiting reagent problem that starts with grams.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
Well, let's do some calculations and see what we get!
First, we need to find the number of moles of barium peroxide and hydrochloric acid we have. We can use their molar masses to do this:
Molar mass of BaO2 = 169.333 g/mol
Molar mass of HCl = 36.461 g/mol
Number of moles of BaO2 = 15.0g / 169.333 g/mol = 0.0886 mol
Number of moles of HCl = (25.0 mL * 0.0272 g/mL) / 36.461 g/mol = 0.0186 mol
According to the balanced equation, the ratio of barium peroxide to hydrogen peroxide is 1:1. So we can expect to get 0.0886 mol of hydrogen peroxide.
Now, to find the mass of hydrogen peroxide, we can multiply the number of moles by its molar mass:
Molar mass of H2O2 = 34.014 g/mol
Mass of hydrogen peroxide = 0.0886 mol * 34.014 g/mol = 3.025 g
So, we should get 3.025 g of hydrogen peroxide as a result.
To find the mass of each reagent left unreacted, we can subtract the number of moles used in the reaction from the total number of moles we started with:
Mass of barium peroxide left unreacted = (15.0g / 169.333 g/mol) - 0.0886 mol = 0.0542 g
Mass of hydrochloric acid left unreacted = (25.0 mL * 0.0272 g/mL) - (0.0186 mol * 36.461 g/mol) = 0.0624 g
So, 0.0542 g of barium peroxide and 0.0624 g of hydrochloric acid will be left unreacted.
Hope that helps! If not, I can always offer you a joke instead.
To solve this problem, we first need to determine the limiting reactant, which is the reactant that will be completely consumed and determines the amount of product formed.
Step 1: Calculate the number of moles of barium peroxide (BaO2):
Mass of BaO2 = 15.0 g
Molar mass of BaO2 = 169.34 g/mol
Number of moles of BaO2 = mass / molar mass = 15.0 g / 169.34 g/mol = 0.0886 mol
Step 2: Determine the number of moles of HCl:
Volume of HCl solution = 25.0 mL
Mass of HCl per mL = 0.0272 g/mL
Mass of HCl = volume x mass per mL = 25.0 mL x 0.0272 g/mL = 0.68 g
Molar mass of HCl = 36.46 g/mol
Number of moles of HCl = mass / molar mass = 0.68 g / 36.46 g/mol = 0.0186 mol
Step 3: Use the balanced equation to find the mole ratio between BaO2 and H2O2:
From the balanced equation, 1 mol of BaO2 reacts to form 1 mol of H2O2.
Step 4: Determine the limiting reactant:
Comparing the moles of BaO2 (0.0886 mol) and HCl (0.0186 mol), we can see that HCl is the limiting reactant because it will be completely consumed before the BaO2.
Step 5: Calculate the mass of hydrogen peroxide produced:
From the balanced equation, the mole ratio between BaO2 and H2O2 is 1:1.
Since the limiting reactant is HCl, the number of moles of H2O2 formed will be equal to the number of moles of HCl reacted.
Number of moles of H2O2 = 0.0186 mol
Molar mass of H2O2 = 34.01 g/mol
Mass of H2O2 = number of moles x molar mass = 0.0186 mol x 34.01 g/mol = 0.6312 g
Therefore, 0.6312 grams of hydrogen peroxide will be produced.
Step 6: Calculate the mass of unreacted reagents:
(a) Mass of unreacted BaO2:
Number of moles of unreacted BaO2 = number of moles of BaO2 initially - number of moles of BaO2 reacted
= 0.0886 mol - 0.0186 mol = 0.0700 mol
Mass of unreacted BaO2 = number of moles x molar mass = 0.0700 mol x 169.34 g/mol = 11.84 g
(b) Mass of unreacted HCl:
Number of moles of unreacted HCl = number of moles of HCl initially - number of moles of HCl reacted
=0.0186 mol - 0.0186 mol = 0 mol (since all of the HCl is reacted)
Mass of unreacted HCl = 0 g
Therefore, the mass of unreacted BaO2 is 11.84 grams, and there is no unreacted HCl left.
Hope this helps!
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.