University Science Books Map ald McQuarrie. Peter A. Rock Ethan Gallogly present

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University Science Books Map ald McQuarrie. Peter A. Rock Ethan Gallogly presented by Sapling Learning oxidizing agent. A 9.63-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 oxidized by 49.7 mL of a 0.110 M aqueous solution of KBrog(aq). The unbalanced equation for the reaction is Bro, (aq)+ sbat (aq) Br (aq) sbs (aq) unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore Number Number 8 view solution Check Answer Next Exit Give Up

Explanation / Answer

Br03- --> Br-

balance the oxygen using H20

BrO3- ---> Br- + 3H20

now balance the Hydrogen atoms using H+

so

Br03- + 6H+ ---> Br- + 3H20

now

balance the charge using electrons

so

Br03- + 6H+ + 6e- ---> Br- + 3H20

now

Sb+3 ---> Sb+5 + 2e-

so

the overall balanced reaction is

Br03- + 6H+ + 3 Sb+3 ---> Br- + 3H20 + 3 Sb+5

we know that

moles = molarity x volume (L)

so

moles of Br03- = 0.11 x 49.7 x 10-3

moles of Br03- = 5.467 x 10-3

now

we can see that

moles of Sb+3 = 3 x moles of Br03-

so

moles of Sb+3 = 3 x 5.467 x 10-3

moles of Sb+3 = 16.401 x 10-3

now

mass = moles x molar mass

so

mass of Sb+3 = 16.401 x 10-3 x 121.76

mass of Sb+3 = 1.997

so

amount of antimony in the sample is 1.997 g

now

% = 1.997 x 100 / 9.63

% = 20.737

so

% antimony in the ore is 20.737 %

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