Answer:
20.66 % of the ore is antimony
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
