Question

The quantity of antimony in a sample can be determined by an oxidation–reduction titration with an oxidizing agent.

Answer 1

Answer:

Explanation:

Step 1: Data given

Mass of stibnite (Sb2S3) = 5.85 grams

The Sb3+(aq) is completely oxidized by 26.6 mL of a  0.125 M aqueous solution of KBrO3(aq).

Step 2: The balanced equation

BrO3-(aq)+ 3Sb^3+(aq) + 6 H+ → Br-(aq) + 3Sb^5+(aq) + 3H2O (l)

Step 3: Calculate moles KBrO3

Moles KBrO3 = molarity * volume

Moles KBrO3 = 0.125 M *0.0266 L

Moles KBrO3 = 0.003325 moles

Step 4: Calculate moles Bro3-

in 1 mol KBrO3 we have 1 mol K+ and 1 mol BrO3-

In 0.003325 moles KBrO3 we have 0.003325 moles BrO3-

Step 5: Calculate moles Sb

For 1 mol BrO3- we need 3 mol Sb^3+ to produce 1 mol Br- and 3 mol Sb^5+

For 0.029085 moles BrO3- we need 3*0.003325 = 0.009975 moles Sb

Step 6: Calculate mass Sb

Mass Sb = moles Sb * molar mass Sb

Mass Sb = 0.009975 moles * 121.76 g/mol

Mass Sb = 1.21 grams

Step 7: Calculate the percentage of Sb in the ore

% Sb = (mass Sb / total mass) * 100%

% Sb = (1.21 grams / 5.85 grams) * 100 %

% Sb = 20.76 %

20.76 % of the ore is antimony

Answer 2

Answer:

The percentage is   k  $= 20.8$%

Explanation:

From the question we are told that

    The mass of the stibnite is  $m_s = 5.86 \ g$

   The volume of   KBrO3(aq) is  $V = 26.6 mL = 26.6 *10^{-3} \ L$

     The concentration  of   KBrO3(aq) is  $C = 0.125 M$

Now the balanced ionic  equation for this reaction is

        $BrO_3 ^{-}+ 3Sb^{3+} + 6H^{+} \to Br^{1-} + 3Sb^{5+} + 3H_2O$

The number of moles of   $BrO_3 ^{-}$ is  

     $n = C *V$

substituting values

     $n = 26.6*10^{-3} * 0.125$

     $n = 0.003325 \ mols$

from the reaction we see that 1 mole of $BrO_3 ^{-}$  reacts with 3 moles of  $Sb^{3+}$

so 0.003325 moles will react with x moles of  $Sb^{3+}$

Therefore

               $x = \frac{0.003325 * 3}{1}$

              $x = 0.009975 \ mols$

Now the molar mass of $Sb^{3+}$ is a constant with a values of  $Z = 121.76 \ g/mol$

Generally the mass of  $Sb^{3+}$ is mathematically represented as

        $m = x * Z$

substituting values

        $m = 1.215 \ g$

The percentage of  Sb(antimony) in the overall mass of the stibnite is mathematically evaluated as

            k  $= \frac{1.215}{5.85 } * 100$

           k  $= 20.8$%