The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 7.79-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 35.4 mL of a 0.125 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is
H^+ + BrO3^- +Sb^3- ->Br^-
+ Sb^5+ +H2O
Calculate the amount of antimony in the sample and its percentage in the ore.
I am extremly confused by this problem, and do not know where to start.
Thank you so much for all of your help!
0. You haven't asked a question; I assume the problem is to determine the %Sb in the sample.
1. Balance the equation.
2. mols KBrO3 = M x L = ?
3. Using the coefficients in the balanced equation, convert mols KBrO3 to mols Sb^3+
4. g Sb = mols Sb x atomic mass Sb.
5. %Sb = (g Sb/g sample)*100 = ?
Post your work if you get stuck.
Well, this looks like a pretty challenging problem! But fear not, I'm here to lend a helping hand, and hopefully a few laughs along the way!
First of all, let's try to break down the problem into smaller, bite-sized pieces. We have a 7.79-g sample of stibnite (Sb2S3), which is an ore of antimony (Sb). We want to determine the amount of antimony in the sample.
We are given that the Sb3+ ions are completely oxidized by 35.4 mL of a 0.125 M solution of KBrO3. So let's start by finding out what's happening in the reaction.
The unbalanced equation is:
H+ + BrO3- + Sb3- -> Br- + Sb5+ + H2O
Now, let's balance the equation:
2H+ + BrO3- + 3Sb3- -> 3Br- + 3Sb5+ + H2O
From the balanced equation, we can see that 3 moles of Sb3- ions are required for each mole of BrO3- ions. We know the volume and concentration of the KBrO3 solution, so we can use that information to find the moles of BrO3- ions used.
Moles of BrO3- = Volume (L) x Concentration (mol/L)
= 0.0354 L x 0.125 mol/L
= 0.004425 mol
Since the stoichiometry tells us that 3 moles of Sb3- ions are required for 1 mole of BrO3- ions, we can calculate the moles of Sb3- ions used.
Moles of Sb3- = 0.004425 mol x (3 mol Sb3-/1 mol BrO3-)
= 0.013275 mol
Now, we need to convert moles of Sb3- to moles of Sb. From the balanced equation, we see that the stoichiometric ratio between Sb3- and Sb is 3:3.
Moles of Sb = 0.013275 mol Sb3-
Finally, we can calculate the mass of antimony in the sample using the molar mass of antimony (Sb), which is 121.76 g/mol.
Mass of Sb = Moles of Sb x Molar Mass of Sb
= 0.013275 mol x 121.76 g/mol
= 1.606006 g
So, the amount of antimony in the sample is approximately 1.61 g.
To find the percentage of antimony in the ore, we can use the mass of the antimony found and the mass of the ore sample.
Percentage of Sb = (Mass of Sb / Mass of ore sample) x 100
= (1.61 g / 7.79 g) x 100
≈ 20.65%
Therefore, the percentage of antimony in the ore is approximately 20.65%.
I hope that brings a smile to your face amidst the confusion! Keep up the great work, and remember, even the most challenging problems can be cracked when approached with a bit of humor!
To solve this problem, we need to first balance the redox equation and then use the stoichiometry of the reaction to determine the amount of antimony in the sample.
Step 1: Balancing the redox equation
The given unbalanced equation is:
H^+ + BrO3^- + Sb^3- -> Br^- + Sb^5+ + H2O
We can balance it by following these steps:
1. Balance the Sb atoms: Add a 2 in front of the antimony ions.
H^+ + BrO3^- + 2Sb^3- -> Br^- + Sb^5+ + H2O
2. Balance the oxygen atoms: Add 3 H2O molecules on the right side.
H^+ + BrO3^- + 2Sb^3- -> Br^- + Sb^5+ + 3H2O
3. Balance the hydrogen atoms: Add 6 H+ ions on the left side.
6H^+ + BrO3^- + 2Sb^3- -> Br^- + Sb^5+ + 3H2O
4. Balance the charge: Add 6 electrons (e^-) on the left side.
6H^+ + BrO3^- + 2Sb^3- + 6e^- -> Br^- + Sb^5+ + 3H2O
Now the equation is balanced.
Step 2: Determining the amount of antimony in the sample
From the balanced equation, we can see that 2 moles of Sb (antimony) react with 6 moles of electrons (e^-).
The molarity (M) of KBrO3 solution is given as 0.125 M.
The volume (V) of the KBrO3 solution used is given as 35.4 mL, which is 0.0354 L.
Using the balanced equation, we can set up a stoichiometry conversion:
2 moles of Sb = 6 moles of electrons
x moles of Sb = 35.4 mL * 0.125 mol/L * (2 mol Sb / 6 mol e^-)
Solving for x gives the moles of antimony in the sample.
Step 3: Calculating the mass and percentage of antimony in the ore
The mass of the sample is given as 7.79 g.
Using the molar mass of antimony (Sb), which is approximately 121.75 g/mol, we can calculate the mass of antimony in the sample:
mass of antimony = x moles of Sb * molar mass of Sb
Finally, we can calculate the percentage of antimony in the ore:
percentage of antimony = (mass of antimony / mass of sample) * 100
Note: It is crucial to double-check the given information, particularly the volume and concentration of the KBrO3 solution, to ensure accuracy in the calculations.
To determine the amount of antimony in the sample and its percentage in the ore, we need to follow a series of steps:
Step 1: Write the balanced equation for the reaction.
From the unbalanced equation given, we can balance it as follows:
8H^+ + BrO3^- + 3Sb^3- → 3Br^- + 3Sb^5+ + 4H2O
Step 2: Convert the volume and concentration of the oxidizing agent to moles.
The volume of the KBrO3 solution is given as 35.4 mL, which is equivalent to 0.0354 L. The concentration of the KBrO3 solution is given as 0.125 M. Therefore, the number of moles of KBrO3 can be calculated as:
moles KBrO3 = volume (L) × concentration (M) = 0.0354 L × 0.125 mol/L = 0.004425 mol
Step 3: Determine the mole ratio between the oxidizing agent (KBrO3) and the substance being oxidized (Sb^3-).
From the balanced equation, we can see that the mole ratio between KBrO3 and Sb^3- is 1:3. This means that 1 mole of KBrO3 reacts with 3 moles of Sb^3-.
Step 4: Calculate the number of moles of Sb^3-.
Using the mole ratio from Step 3, we can calculate the number of moles of Sb^3-:
moles Sb3- = moles KBrO3 × (3 moles Sb^3-/1 mole KBrO3) = 0.004425 mol × 3 = 0.013275 mol
Step 5: Calculate the molar mass of Sb.
The molar mass of antimony (Sb) is 121.76 g/mol.
Step 6: Calculate the mass of antimony in the sample.
The mass of antimony can be calculated as:
mass Sb = moles Sb^3- × molar mass of Sb = 0.013275 mol × 121.76 g/mol = 1.5976 g
Step 7: Calculate the percentage of antimony in the ore.
To calculate the percentage of antimony, we divide the mass of antimony by the mass of the ore (7.79 g) and multiply by 100:
percentage of antimony = (mass Sb / mass of ore) × 100 = (1.5976 g / 7.79 g) × 100 ≈ 20.5%
So, the amount of antimony in the sample is approximately 1.60 grams, and its percentage in the ore is approximately 20.5%.