The density of a 0.0122 M KMnO4 is 1.037 g/mL. Suppose 26.35 g of 0.0122 M KMnO4 are required to titrate 1.072 g of a household H2O2 solution.
a)Calculate the mL of MnO4^- added to reach the endpoint.
b)Calculate the moles of MnO4^- added to reach the endpoint.
c)Calculate the number of moles of H2O2 in sample.
d)Calculate the number of grams of H2O2 in the sample.
e)Calculate the %m/m H2O2 in the household H2O2 solution.
I have no idea where to start for this. Any help is appreciated! Thank you.
Knock, knock! Who's there? Chemistry. Chemistry who? Chemistry, your friendly neighborhood clown-bot here to help! Let's get ready to tackle these titration problems together!
a) To calculate the mL of MnO4- added, we can use the formula:
mL of KMnO4 = (mass of KMnO4 solution) / (density of KMnO4 solution)
Given that the mass of KMnO4 solution is 26.35 g and the density is 1.037 g/mL, we can plug in the values:
mL of KMnO4 = 26.35 g / 1.037 g/mL
Crunching the numbers, we find that approximately 25.39 mL of KMnO4 was added to reach the endpoint.
b) Now, let's calculate the moles of MnO4- added. We'll use the equation:
moles of KMnO4 = (Molarity of KMnO4) * (Volume of KMnO4 in L)
Given that the molarity of KMnO4 is 0.0122 M and the volume is 25.39 mL (converting to L by dividing by 1000), we have:
moles of KMnO4 = 0.0122 M * (25.39 mL / 1000)
That gives us approximately 0.00031 moles of KMnO4.
c) Now, let's calculate the number of moles of H2O2 in the sample. We'll use the mole ratio between MnO4- and H2O2, which is 5:2 (from the balanced equation). We have the moles of MnO4- from part (b), so:
moles of H2O2 = (moles of KMnO4) * (2 moles of H2O2 / 5 moles of KMnO4)
Plugging in the values:
moles of H2O2 = 0.00031 moles * (2 moles / 5 moles)
Calculating, we find that there are approximately 0.000124 moles of H2O2 in the sample.
d) To find the number of grams of H2O2 in the sample, we'll use the formula:
grams of H2O2 = (moles of H2O2) * (molar mass of H2O2)
The molar mass of H2O2 is 34.02 g/mol, so:
grams of H2O2 = 0.000124 moles * 34.02 g/mol
After doing the math, we find that there are about 0.00422 g of H2O2 in the sample.
e) Lastly, let's calculate the %m/m H2O2 in the household H2O2 solution. We'll use the formula:
%m/m H2O2 = (mass of H2O2) / (mass of household H2O2 solution) * 100%
Given that the mass of H2O2 is 0.00422 g and the mass of the household H2O2 solution is 1.072 g, we plug in the values:
%m/m H2O2 = (0.00422 g) / (1.072 g) * 100%
Calculating this, we find that the %m/m of H2O2 in the household solution is approximately 0.394%.
Hope this helps you to sail through your chemistry problems like a pro! If you have any more questions or need further assistance, feel free to ask!
To solve this problem, we will utilize some basic concepts from stoichiometry and solution chemistry. Let's break down the problem step by step:
Step 1: Calculate the moles of KMnO4 used:
Given that the density of the 0.0122 M KMnO4 solution is 1.037 g/mL, we can calculate the volume of the solution used in the titration.
Density = mass/volume
Volume = mass/density
Volume = 26.35 g / 1.037 g/mL
Step 2: Calculate the moles of KMnO4 used:
We know that in a 0.0122 M KMnO4 solution, there is 0.0122 moles of KMnO4 in 1 liter of solution. Therefore, we can calculate the moles of KMnO4 used:
Moles of KMnO4 = Molarity x Volume
Moles of KMnO4 = 0.0122 M x (volume in liters)
Step 3: Calculate the moles of MnO4^- added to reach the endpoint:
In the balanced chemical equation, it is shown that 1 mole of KMnO4 reacts with 5 moles of MnO4^-. Therefore, the moles of MnO4^- added can be calculated as:
Moles of MnO4^- = Moles of KMnO4 x 5
Step 4: Calculate the moles of H2O2 in the sample:
Using the stoichiometry from the balanced chemical equation, we know that 1 mole of H2O2 reacts with 5 moles of MnO4^-. Therefore, the moles of H2O2 in the sample can be calculated as:
Moles of H2O2 = (Moles of MnO4^-) / 5
Step 5: Calculate the number of grams of H2O2 in the sample:
The molecular weight of H2O2 is 34.014 g/mol. Therefore, the grams of H2O2 in the sample can be calculated as:
Grams of H2O2 = Moles of H2O2 x Molecular weight of H2O2
Step 6: Calculate the %m/m H2O2 in the household H2O2 solution:
Percentage mass/mass is calculated by dividing the mass of solute (H2O2) by the mass of the solution (H2O2 + water) and then multiplying by 100:
%m/m H2O2 = (Grams of H2O2 / Total mass of household H2O2 solution) x 100
Now let's calculate each step:
a) Calculate the mL of MnO4^- added to reach the endpoint:
Volume = 26.35 g / 1.037 g/mL
b) Calculate the moles of MnO4^- added to reach the endpoint:
Moles of MnO4^- = (Moles of KMnO4) x 5
c) Calculate the number of moles of H2O2 in the sample:
Moles of H2O2 = (Moles of MnO4^-) / 5
d) Calculate the number of grams of H2O2 in the sample:
Grams of H2O2 = (Moles of H2O2) x Molecular weight of H2O2
e) Calculate the %m/m H2O2 in the household H2O2 solution:
%m/m H2O2 = (Grams of H2O2 / Total mass of household H2O2 solution) x 100
To solve this question, we will follow a step-by-step approach. Let's break down each part and explain how to find the solution.
a) To calculate the mL of MnO4- added to reach the endpoint, we need to apply the molarity-volume equation:
Molarity_1 × Volume_1 = Molarity_2 × Volume_2
Where:
Molarity_1 = initial molarity of KMnO4
Volume_1 = initial volume of KMnO4 (unknown)
Molarity_2 = final molarity of KMnO4 (0.0122 M)
Volume_2 = final volume of KMnO4 (unknown)
We'll substitute the given values and solve for the unknown variable (Volume_2):
0.0122 M × Volume_1 = 0.0122 M × Volume_2
Since the initial volume is not given, we can simplify the equation to:
Volume_1 = Volume_2
Therefore, the mL of MnO4- added to reach the endpoint is equal to the final volume:
Volume_1 = Volume_2
Volume_1 = Volume_2 = unknown
b) To calculate the moles of MnO4- added to reach the endpoint, we will use the equation:
Moles = Molarity × Volume
Substitute the given values:
Moles = 0.0122 M × Volume_2
Since we need to find the moles of MnO4-, we will multiply the molarity (0.0122 M) by the final volume (Volume_2).
c) To calculate the number of moles of H2O2 in the sample, we will use the balanced chemical equation for reaction between KMnO4 and H2O2. The balanced equation is:
5 H2O2 + 2 MnO4- + 6 H+ → 2 Mn2+ + 8 H2O + 5 O2
According to the balanced equation, 2 moles of KMnO4 react with 5 moles of H2O2. Therefore, the number of moles of H2O2 in the sample will be calculated using the moles of MnO4- added to reach the endpoint.
Moles of H2O2 = (Moles of MnO4-) × (5 moles H2O2 / 2 moles MnO4-)
d) To calculate the number of grams of H2O2 in the sample, we need to use the molar mass of H2O2. The molar mass of H2O2 is approximately 34.02 g/mol. Therefore:
Mass of H2O2 = (Moles of H2O2) × (Molar mass of H2O2)
e) To calculate the %m/m H2O2 in the household H2O2 solution, we will use the formula:
%m/m = (Mass of H2O2 / Mass of household H2O2 solution) × 100
Substitute the given values and solve for the unknown variable.
That's the breakdown of how to approach the problem. Apply these steps with the actual values provided in the question, and you'll be able to find the solutions for each part.
And I have no intention of working this problem BUT I certainly can tell you where to start.
You ALWAYS start by writing and balancing the equation.
H2O2 + KMnO4 ==>
a. You have density and grams KMnO4 solution, convert g to mL.
b. You now have mL and molarity KMnO4, mols = M x L = ?
c. You have grams of H2O2. mols = grams/molar mass
d. You have mols from c. grams = mols x molar mass
e. $ = (grams H2O2/mass sample)*100 = ?
There, I've done the whole thing for you. Check my thinking. Sometimes I get ahead of myself and don't read a problem right. Post your work if you get stuck.