A 0.345g sample of anhydrous BeC2O4, which contains an inert impurity, was dissolved in a sufficient water to produce 100ml of solution. A 20.0ml portion of the solution was titrated w/ KMnO4(aq). The balanced equation for the reaction that occurred is as follows:
16H^+(aq) + 2MnO4^-(aq) + 5C2O4^2-(aq) ---> 2Mn^2+(aq) + 10CO2(g) + 8H2O(l)
the volume of 0.0150M KMnO4(aq) required to reach the equivalence point was 17.80ml.
a) identify the reducing agent in the reaction
b) for the titration at the equivalence point, calculate the number of moles of each of the following that reacted:
MnO4^-(aq)
C2O4^2-(aq)
c) calculate the total # of moles of C2O4^2-(aq) that were present in the 100ml of prepared solution.
d)calculate the mass percent of BeC2O4(s) in the impure 0.345g sample.
can u give me the steps to solving this please? thanx
A 0.345g sample of anhydrous BeC2O4, which contains an inert impurity, was dissolved in a sufficient water to produce 100ml of solution. A 20.0ml portion of the solution was titrated w/ KMnO4(aq). The balanced equation for the reaction that occurred is as follows:
16H^+(aq) + 2MnO4^-(aq) + 5C2O4^2-(aq) ---> 2Mn^2+(aq) + 10CO2(g) + 8H2O(l)
the volume of 0.0150M KMnO4(aq) required to reach the equivalence point was 17.80ml.
a) identify the reducing agent in the reaction is the substance that is oxidized.
b) for the titration at the equivalence point, calculate the number of moles of each of the following that reacted:
MnO4^-(aq) mols=L x M =??
C2O4^2-(aq) = ?? mols MnO4^- *(5 mols C2O4^-2)/2 mols MnO4^-). Use the coefficients in the balanced equation. Note units of mols permangante cancel leaving mols oxalate.
c) calculate the total # of moles of C2O4^2-(aq) that were present in the 100ml of prepared solution. b gives mols present in 20 mL. That was 1/5 the initial sample; therefore, multiply mols found in the titration by 5.
d)calculate the mass percent of BeC2O4(s) in the impure 0.345g sample.
mols x molar mass = grams.
%= (grams/weight sample)x100
are these right::
a) Mn
b) 3x10-4 and 7.5x10-4
c) 1.125x10-6
d) ??
i. (C2O4)-2 Loss of electrons....
ii. (MnO4)-1:
0.0150M*.0178L = mol(MnO4)-1
=>2.67*10^-4 mol
(C2O4)-2:
2.67*10^-4 mol * 5(C2O4)/2(MnO4) (ratio)
=>6.68*10^-4 mol
iii. 6.68*10^-4 mol * 5 (100.ml/20.ml=5)
=>3.34*10^-3 mol
iv. 3.34*10^-3 mol * 97.032gfm = 0.324g
(0.324g / 0.345g) * 100.
=>93.9%
a) Mn is not the reducing agent in this reaction. In this reaction, the reducing agent is C2O4^2-(aq) because it loses electrons and is oxidized to CO2(g).
b) To calculate the number of moles of MnO4^-(aq) and C2O4^2-(aq) that reacted at the equivalence point:
Moles of MnO4^-(aq) = (volume of KMnO4(aq) × molarity of KMnO4(aq)) = (17.80 mL × 0.0150 M) = 0.267 mol
Moles of C2O4^2-(aq) = (moles of MnO4^-(aq) × 5 mol C2O4^2-/2 mol MnO4^-) = (0.267 mol × 5 mol C2O4^2-/2 mol MnO4^-) = 0.667 mol
c) The number of moles of C2O4^2-(aq) that were present in the 100 mL of prepared solution can be calculated by multiplying the moles found in the titration (0.667 mol) by the dilution factor, which is 5:
Total moles of C2O4^2-(aq) = (moles of C2O4^2-(aq) × dilution factor) = (0.667 mol × 5) = 3.335 mol
d) To calculate the mass percent of BeC2O4(s) in the impure 0.345 g sample:
First, calculate the molar mass of BeC2O4:
Molar mass of BeC2O4 = (1 × atomic mass of Be) + (2 × atomic mass of C) + (4 × atomic mass of O)
= (1 × 9.012 g/mol) + (2 × 12.011 g/mol) + (4 × 16.00 g/mol)
= 47.014 g/mol
Next, calculate the moles of BeC2O4:
Moles of BeC2O4 = (mass of sample / molar mass of BeC2O4) = (0.345 g / 47.014 g/mol) = 0.00734 mol
Finally, calculate the mass percent:
Mass percent = (mass of BeC2O4 / weight of sample) × 100
= (0.00734 mol × 47.014 g/mol / 0.345 g) × 100
= 10.03%
So, the mass percent of BeC2O4(s) in the impure 0.345 g sample is approximately 10.03%.
Yes, your answers for parts (a) and (b) are correct. Let's continue with part (c) and (d).
c) To calculate the total number of moles of C2O4^2-(aq) present in the 100 mL of the prepared solution, we can use the ratio obtained from the balanced equation:
5 moles C2O4^2- : 2 moles MnO4^-
We already determined that 7.5x10^(-4) moles of MnO4^- reacted in the titration. Since this is the molar ratio between MnO4^- and C2O4^2-, we can multiply it by 5 to obtain the number of moles of C2O4^2-:
7.5x10^(-4) mol MnO4^- * 5 mol C2O4^2-/2 mol MnO4^- = 1.875x10^(-3) mol C2O4^2-
Therefore, there were 1.875x10^(-3) moles of C2O4^2- in the 100 mL of prepared solution.
d) To calculate the mass percent of BeC2O4(s) in the impure 0.345 g sample, we need to first calculate the mass of BeC2O4(s) present in the sample. We already know the number of moles of C2O4^2- (1.875x10^(-3) mol) from part (c). We can use the molar ratio obtained from the balanced equation to determine the number of moles of BeC2O4:
5 moles C2O4^2- : 1 mole BeC2O4
1.875x10^(-3) mol C2O4^2- * 1 mol BeC2O4/5 mol C2O4^2- = 3.75x10^(-4) mol BeC2O4
Now, we can calculate the mass of BeC2O4:
3.75x10^(-4) mol BeC2O4 * (2 mol Be + 1 mol C + 4 mol O) * molar mass BeC2O4
= 3.75x10^(-4) mol * (2(9.012 g/mol) + 12.01 g/mol + 4(16.00 g/mol))
= 3.75x10^(-4) mol * 57.04 g/mol
= 2.135 g BeC2O4
Finally, we can calculate the mass percent of BeC2O4 in the impure 0.345 g sample:
(mass of BeC2O4/mass of sample) * 100
= (2.135 g/0.345 g) * 100
= 618.84%
The mass percent of BeC2O4 in the impure 0.345 g sample is approximately 618.84%.