Question

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 5.13 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 27.7 mL of a .105 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is:_____________.

BrO-3 (aq) + Sb3+(aq)--> Br-(aq)+ Sb5+(aq (unbalanced)
Calculate the amount of antimony in the sample (grams) and its percentage in the ore (%)

Answer 1

The amount of antimony is 1.062 g and its percentage is 20.7%

Step-by-step explanation:

The reaction is:

3Sb3+ + BrO3- + 6H+ = 3Sb+5 + Br + 3H2O

The moles of KBrO3 is:

`nKBrO3=27.7*0.105=2.9085mmoles`

The moles of Sb3+ is:

`nSb3+=3*2.9085=8.7255mmoles`

The mass of Sb3+ is:

`massSb3+=8.7255x10^(-3) *121.76=1.062g`

The percentage is:

`P=(1.062)/(5.13) *100=20.7`%

Answer 2

20.66 % of the ore is antimony

Step-by-step explanation:

Step 1: Data given

Mass of stibnite (Sb2S3) = 5.13 grams

The Sb3+(aq) is completely oxidized by 27.7 mL of a 0.105 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 = 0.105 M *0.0277 L

Moles KBrO3 = 0.0029085 moles

Step 4: Calculate moles Bro3-

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

In 0.0029085 moles KBrO3 we have 0.0029085 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*0.0029085 = 0.0087255 moles Sb

Step 6: Calculate mass Sb

Mass Sb = moles Sb * molar mass Sb

Mass Sb = 0.0087255 moles * 121.76 g/mol

Mass Sb = 1.06 grams

Step 7: Calculate the percentage of Sb in the ore

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

% Sb = (1.06 grams / 5.13 grams) * 100 %

% Sb = 20.66 %

20.66 % of the ore is antimony