Question

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

A 5.85 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 26.6 mL of a 0.125 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is:
BrO3-(aq) + Sb3+(aq) ------> Br-(aq) + Sb5+(aq) (unbalanced)
(A) Calculate the amount of antimony in the sample and its percentage in the ore.

Answer 1

Step-by-step 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

The percentage is k
`= 20.8`%

Step-by-step 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 = (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
`= (1.215)/(5.85 ) * 100`

k
`= 20.8`%