What mass of 3.0 wt% H2O2 solution is required to provide a 25% excess of reagent for the following reaction with 12 dietary iron tablets that contain ~15 mg of iron per tablet?
To calculate the mass of 3.0 wt% H2O2 solution needed, we need to determine the amount of H2O2 required to provide a 25% excess for the reaction with the iron tablets.
First, let's calculate the total amount of iron in the 12 dietary iron tablets:
12 tablets x 15 mg iron per tablet = 180 mg of iron
Next, we need to determine the stoichiometry of the reaction between H2O2 and iron. The balanced reaction equation is:
2 H2O2 + 2 Fe -> 2 Fe(OH)2
From the balanced equation, we can see that 2 moles of H2O2 react with 2 moles of Fe.
Now, let's calculate the molar amount of iron in the 180 mg:
Molar mass of Fe = 55.85 g/mol
Molar amount of iron = (180 mg / 1000) / 55.85 g/mol = 0.0032158 mol
To provide a 25% excess of H2O2, we need to determine the molar amount of H2O2 required:
Molar amount of H2O2 = 0.0032158 mol x 1.25 = 0.00401975 mol
To calculate the mass of H2O2 needed, we need to convert the molar amount to mass using the concentration of the 3.0 wt% H2O2 solution.
The concentration of 3.0 wt% H2O2 means that there are 3.0 g of H2O2 per 100 g of solution. This can also be expressed as 3.0 g/100 mL.
First, let's calculate the volume of the 3.0 wt% H2O2 solution needed in mL:
Molar amount of H2O2 = Mass of H2O2 / Molar mass of H2O2
Mass of H2O2 = Molar amount of H2O2 x Molar mass of H2O2 = 0.00401975 mol x (34.02 g/mol) = 0.1365 g
Let's assume the density of the 3.0 wt% H2O2 solution is approximately 1 g/mL. Therefore, the mass and volume of the solution needed are the same.
Mass and volume of 3.0 wt% H2O2 solution needed = 0.1365 g = 0.1365 mL
Therefore, approximately 0.1365 g or 0.1365 mL of the 3.0 wt% H2O2 solution is required to provide a 25% excess of reagent for the reaction with 12 dietary iron tablets.
To determine the mass of 3.0 wt% H2O2 solution required, we will need to calculate the moles of H2O2 required for the given reaction and then convert it back to mass using the weight percentage provided.
Let's start by writing the balanced chemical equation for the reaction:
2H2O2 + 2Fe2+ -> 2Fe3+ + 2H2O + O2
From the equation, we see that 2 moles of H2O2 are required per 2 moles of Fe2+.
Since there are 12 dietary iron tablets with approximately 15 mg of iron per tablet, the total mass of iron is: 12 tablets * 15 mg/tablet = 180 mg = 0.18 g.
Now, let's calculate the moles of Fe2+ required:
Moles of Fe2+ = (mass of Fe) / (molar mass of Fe)
Molar mass of Fe = 55.845 g/mol (atomic mass of iron)
Moles of Fe2+ = 0.18 g / 55.845 g/mol ≈ 0.00322 mol
Since the reaction requires 2 moles of H2O2 for 2 moles of Fe2+, we need the same number of moles of H2O2 for the reaction.
Now, let's calculate the mass of H2O2 required:
Mass of H2O2 = Moles of H2O2 * Molar mass of H2O2
Molar mass of H2O2 = 34.0147 g/mol (oxygen atomic mass + 2 hydrogen atomic mass)
Moles of H2O2 = 0.00322 mol
Mass of H2O2 = 0.00322 mol * 34.0147 g/mol ≈ 0.110 g
Since the given solution is a 3.0 wt% H2O2 solution, we need to find the mass of the solution that corresponds to 0.110 g of H2O2.
Weight percentage = (mass of solute / mass of solution) * 100%
Rearranging the equation, we get:
Mass of solute = (weight percentage / 100%) * Mass of solution
In this case, we have a 3.0 wt% H2O2 solution and we want to find the mass of the solution. Assuming a solution mass of 100 g, we can calculate the mass of H2O2 present in the solution:
Mass of solute (H2O2) = (3.0 / 100%) * Mass of solution
Rearranging the equation, we can solve for the mass of the solution:
Mass of solution = Mass of solute (H2O2) / (3.0 / 100%)
Mass of solution = 0.110 g / (3.0 / 100%) = 3.67 g
Therefore, approximately 3.67 grams of the 3.0 wt% H2O2 solution is required to provide a 25% excess of reagent for the given reaction.