The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 6.35-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCl(aq) and passed over a reducing agent so that all the antimony is in the form Sb3 (aq). The Sb3 (aq) is completely oxidized by 27.7 mL of a 0.130 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

BrO3^- + Sb^3+= Br^- + Sb^5+

Calculate the amount of antimony in the sample and its percentage in the ore.

BrO^3- + 3Sb^3+ + 6H^+ -----> Br^- + 3Sb^5+ +3H2o

Balance the equation.

moles BrO3^- = M x L = ?
Use the coefficients in the balanced equation to convert moles BrO3^- to moles Sb.
Then grams Sb = mols x molar mass
%Sb = (grams Sb/mass sample)*100 = ?

Well, I gotta say, this question is quite a titration sensation! Let's break it down with some clown flair.

First, we need to balance that equation. It seems like BrO3^- and Sb^3+ are having a cheerful exchange, like a chemistry dance-off. They shake things up and transform into Br^- and Sb^5+. So, let's make sure this equation is ready to hit the dance floor:

2BrO3^- + 3Sb^3+ → 3Br^- + 3Sb^5+

Now, we know the stoichiometry of the reaction, and you've given us the molarity and volume of KBrO3(aq) needed to oxidize the antimony. So, we can calculate the moles of Sb^3+ in our reaction.

Moles of Sb^3+ = (0.130 M) * (0.0277 L)

Next, we need to convert the moles of Sb^3+ to moles of Sb. Since the stoichiometry between Sb^3+ and Sb is 1:1, there's no need for any clown math here.

Moles of Sb = Moles of Sb^3+

Now, let's calculate the grams of Sb in the sample by using the molar mass of antimony, which is about 121.76 g/mol.

Grams of Sb = Moles of Sb * (121.76 g/mol)

Finally, we can calculate the percentage of antimony in the ore by dividing the grams of Sb by the mass of the sample and multiplying by 100.

Percentage of Sb = (Grams of Sb / Mass of Sample) * 100

And there you have it, the amount of antimony in the sample and its percentage in the ore! I hope this answer made you smile as much as I did while crunching those numbers.

To calculate the amount of antimony in the sample and its percentage in the ore, we need to consider the stoichiometry of the reaction and the volume and concentration of the oxidizing agent used.

1. Determine the moles of oxidizing agent used:
Moles of KBrO3 = volume (L) x concentration (mol/L)
Moles of KBrO3 = 0.0277 L x 0.130 mol/L
Moles of KBrO3 = 0.003601 mol

2. Using the unbalanced equation, determine the molar ratio between the oxidizing agent (KBrO3) and the antimony (Sb):
1 mol KBrO3: 1 mol Sb^3+

3. Calculate the moles of antimony (Sb) oxidized:
Moles of Sb^3+ = 0.003601 mol

4. Since all the antimony in the 6.35 g sample is in the form of Sb^3+, we can calculate the molar mass of antimony from its atomic mass:
Molar mass of antimony (Sb) = 121.76 g/mol

5. Calculate the amount of antimony in the sample:
Amount of Sb = Moles of Sb^3+ x Molar mass of Sb
Amount of Sb = 0.003601 mol x 121.76 g/mol
Amount of Sb = 0.438 g

6. Calculate the percentage of antimony in the ore:
Percentage of Sb = (Amount of Sb / Sample mass) x 100%
Percentage of Sb = (0.438 g / 6.35 g) x 100%
Percentage of Sb = 6.90%

Therefore, the amount of antimony in the sample is 0.438 g, and its percentage in the ore is 6.90%.

To calculate the amount of antimony in the sample and its percentage in the ore, we need to determine the number of moles of antimony reacted with the oxidizing agent (KBrO3).

Step 1: Balanced Equation
Given the unbalanced equation: BrO3^- + Sb^3+= Br^- + Sb^5+
We need to balance it. It can be balanced as follows:
2BrO3^- + 3Sb^3+= 2Br^- + 3Sb^5+

Step 2: Stoichiometry
From the balanced equation, we can see that the stoichiometric ratio between BrO3^- and Sb^3+ is 2:3. Therefore, 2 moles of BrO3^- are required for every 3 moles of Sb^3+.

Step 3: Calculation
We are given that the volume of KBrO3 solution used is 27.7 mL, and its concentration is 0.130 M. We can use this information to calculate the moles of KBrO3 used:

Moles of KBrO3 = Volume (L) x Concentration (mol/L)
= 27.7 mL / 1000 mL/L x 0.130 mol/L
= 0.003601 mol

Using the stoichiometry from step 2, we can determine the moles of Sb^3+ reacted:

Moles of Sb^3+ = (3/2) x Moles of KBrO3
= (3/2) x 0.003601 mol
= 0.0054015 mol

Step 4: Molar Mass
To calculate the amount of antimony in the sample and its percentage in the ore, we need to know the molar mass of antimony (Sb). The molar mass of Sb is 121.76 g/mol.

Step 5: Calculate Amount of Antimony
Amount of Antimony = Moles of Sb^3+ x Molar Mass of Sb
= 0.0054015 mol x 121.76 g/mol
= 0.656 g

Step 6: Calculate Percentage of Antimony
The mass of the sample is given as 6.35 g. Therefore, the percentage of antimony in the ore is:

Percentage of Antimony = (Amount of Antimony / Mass of Sample) x 100%
= (0.656 g / 6.35 g) x 100%
= 10.32%

Therefore, the amount of antimony in the sample is approximately 0.656 g, and the percentage of antimony in the ore is approximately 10.32%.