The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. a 6.75-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 36.5 ml of a 0.105 m aqueous solution of kbro3(aq). the unbalanced equation for the reaction is

1.40 g

20.7 %

Explanation:

There is some info missing. I think this is the complete question.

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 6.75-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 36.5 mL of a 0.105 M aqueous solution of KBrO₃(aq). The unbalanced equation for the reaction is

BrO₃⁻(aq) + Sb³⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq) (unbalanced)

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

We will start by balancing the reaction through the ion-electron method. The half-reactions are:

Reduction: BrO₃⁻(aq) → Br⁻(aq)

Oxidation: Sb³⁺(aq) → Sb⁵⁺(aq)

We will perform the mass balance adding H⁺(aq) and H₂O(l) where necessary.

6 H⁺(aq) + BrO₃⁻(aq) → Br⁻(aq) + 3 H₂O(l)

Sb³⁺(aq) → Sb⁵⁺(aq)

Now, we will balance the charges adding electrons where necessary.

6 H⁺(aq) + BrO₃⁻(aq) + 6 e⁻ → Br⁻(aq) + 3 H₂O(l)

Sb³⁺(aq) → Sb⁵⁺(aq) + 2 e⁻

Then, we make sure that the number of electrons gained and lost are the same and add both half-reactions.

6 H⁺(aq) + BrO₃⁻(aq) + 6 e⁻ → Br⁻(aq) + 3 H₂O(l)

3 Sb³⁺(aq) → 3 Sb⁵⁺(aq) + 6 e⁻

6 H⁺(aq) + BrO₃⁻(aq) + 3 Sb³⁺(aq) → Br⁻(aq) + 3 H₂O(l) + 3 Sb⁵⁺(aq)

The moles of KBrO₃ are:

36.5 × 10⁻³ L × 0.105 mol/L = 3.83 × 10⁻³ mol

The moles of BrO₃⁻ are equal to the moles of KBrO₃.

The molar ratio of BrO₃⁻ to Sb³⁺ is 1:3. The moles of Sb³⁺ are 3 × (3.83 × 10⁻³ mol) = 0.0115 mol

The moles of Sb are equal to the moles of Sb³⁺.

The molar mass of Sb is 121.76 g/mol. The amount of Sb in the sample is:

0.0115 mol × (121.76 g/mol) = 1.40 g

The mass percent of Sb in the sample is:

(1.40 g / 6.75 g) × 100% = 20.7%