A 0,518 g sample of limestone was dissolved and then the calcium was precipitated as calcium oxalate ( CaC2O4). Afer filtrating the precipitate, it was needed 40,0 ml of KMnO4 0,250 N acidified with H2SO4 to titrate it. ¿ What is the %w of CaO in the sample? The equation for the titration is:
MnO-4 + CaC2O4 --> CaSO4 + Mn-2 +CO2 + H2O
The equation is not balanced because I have to use equivalents to get the result.
The result has to be 54,2%
I don't know what to do. I would like to know the steps
1 Ca = 1 CaO = 1 C2O4^2-
2MnO4^- + 5C2O4^2 ==> .....-
mew = milliequivalent weight.
mL x N x mew = grams
Then (grams/mass sample)*100 = ?
mL = 40.0
N = 0.250
mew = 1/2(molar mass CaO)
I get answer of 52.1%. The molar mass CaO may be a little different in your text/notes.
All of the numbers are in the problem except for the mew of CaO.
1 mol Ca = 1 mol CaO = 1 mol C2O4^2-
Titration step is
5 mol C2O4^2- = 1 mol MnO4^-
So 5 mol CaO = 2 mol MnO4^-
which makes
5/2 mol CaO = 1 mol MnO4^0
Now you want to change that to equivalents. The MnO4^- changes by 5 electrons so
(5/2)/5 = 1 eq MnO4^-
and that makes equivalent weight CaO 1/2 molar mass. (i.e., (5/2*5)= 1/2.
To calculate the percentage by weight of CaO in the limestone sample, you will need to follow a series of steps. Here's a step-by-step guide on how to solve this problem:
Step 1: Determine the molar mass of CaO
- The molar mass of CaO (calcium oxide) is the sum of the molar masses of calcium (Ca) and oxygen (O).
- Calcium has a molar mass of approximately 40.08 g/mol, and oxygen has a molar mass of approximately 16.00 g/mol.
- Therefore, the molar mass of CaO is 40.08 + 16.00 = 56.08 g/mol.
Step 2: Calculate the molarity of KMnO4 solution
- The molarity (M) of a solution is defined as the moles of solute divided by the volume of the solution in liters.
- The volume of the KMnO4 solution used for titration is given as 40.0 mL, which is equivalent to 0.0400 L.
- The molarity (M) is given as 0.250 N, which means it is 0.250 moles per liter.
- Therefore, the moles of KMnO4 in the solution used for titration is 0.250 mol/L * 0.0400 L = 0.010 mol.
Step 3: Determine the stoichiometry of the reaction
- Based on the given equation for the titration, the stoichiometry of the reaction between KMnO4 and CaC2O4 is 1:1. That means one mole of KMnO4 reacts with one mole of CaC2O4.
Step 4: Calculate the moles of CaC2O4
- Since the stoichiometry is 1:1, the moles of CaC2O4 will be the same as the moles of KMnO4 used in the reaction, which is 0.010 mol.
Step 5: Convert the moles of CaC2O4 to the mass of CaO
- Since CaC2O4 contains one mole of CaO, the mass of CaO can be calculated using the molar mass of CaO (as calculated in Step 1).
- Mass of CaO = Moles of CaC2O4 * Molar mass of CaO
- Mass of CaO = 0.010 mol * 56.08 g/mol = 0.5608 g
Step 6: Calculate the percentage by weight of CaO in the limestone sample
- The percentage by weight is calculated by dividing the mass of CaO by the mass of the limestone sample and then multiplying by 100%.
- Percentage by weight = (Mass of CaO / Mass of limestone sample) * 100%
- Percentage by weight = (0.5608 g / 0.518 g) * 100% = 108.2%
Therefore, the percentage by weight of CaO in the limestone sample is approximately 108.2%. Since this value exceeds 100%, it appears there may be an error in the given information or calculations.
To calculate the percentage of CaO in the sample, you need to determine the number of moles of CaO from the amount of KMnO4 used in the titration and then convert it to a percentage. Here are the steps to follow:
1. Determine the number of moles of KMnO4:
- Convert the volume of KMnO4 used in the titration (40.0 mL) to liters by dividing by 1000: 40.0 mL = 0.040 L.
- Multiply the volume by the molarity of KMnO4 (0.250 N) to get moles: 0.040 L * 0.250 mol/L = 0.0100 mol KMnO4.
2. Write the balanced equation for the reaction:
MnO4- + CaC2O4 -> CaSO4 + Mn2+ + CO2 + H2O
3. Determine the stoichiometric ratio between KMnO4 and CaC2O4:
From the balanced equation, you can see that the stoichiometric ratio is 1:1. It means that 1 mole of KMnO4 reacts with 1 mole of CaC2O4.
4. Calculate the number of moles of CaC2O4:
Since the stoichiometric ratio is 1:1, the number of moles of CaC2O4 will be the same as the number of moles of KMnO4, which is 0.0100 mol.
5. Calculate the molar mass of CaO:
The molar mass of CaO is equal to the atomic mass of calcium (40.08 g/mol) plus the atomic mass of oxygen (16.00 g/mol), which gives a molar mass of 56.08 g/mol.
6. Calculate the mass of CaO:
Multiply the number of moles of CaC2O4 (0.0100 mol) by the molar mass of CaO (56.08 g/mol): 0.0100 mol * 56.08 g/mol = 0.5608 g.
7. Calculate the percent weight of CaO in the sample:
The percent weight is calculated by dividing the mass of CaO (0.5608 g) by the mass of the sample (0.518 g) and multiplying by 100: (0.5608 g / 0.518 g) * 100 = 108.17%.
Note: The result obtained is not the expected 54.2%. Please recheck the data and make sure all the calculations are correctly performed.