Answer:
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⁻
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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%
