0.4225 g copper ore 0.02100 M Na2S2O3 29.96 mL CuCO3

0.4225 g copper ore 0.02100 M Na2S2O3 29.96 mL CuCO3

 

 

A sample of a copper ore with a mass of 0.4225 g was dissolved in acid.

A solution of potassium iodide was added, which caused the reaction:

2Cu^2+(aq) + 5I^-(aq) → I3^-(aq) + 2CuI(s)

 

The I3^- that formed reacted quantitatively

with exactly 29.96 mL of 0.02100 M Na2S2O3

according to the following equation:

I3^-(aq) + 2S2O3^2-(aq) → 3I^-(aq) + S4O6^2-(aq)

 

a) What was the percentage by mass of copper in the ore?

b) If the ore contained CuCO3,

what was the percentage by mass of CuCO3 in the ore?

 

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(0.02100 mol/L) (29.96/1000 L) = 0.00062916 mol Na2S2O3

= 0.00062916 mol S2O3^2-

 

 

 

I3^- : S2O3^2- = 1 : 2 계수비(= 몰수비) 이고,

Cu^2+ : I3^- = 2 : 1 계수비(= 몰수비) 이므로,

 

(0.00062916 mol S2O3^2-) (1 mol I3^- / 2 mol S2O3^2-) (2 mol Cu^2+ / 1 mol I3^-)

= (0.00062916) (1 / 2) (2 / 1)

= 0.00062916 mol Cu^2+

 

 

 

Cu의 몰질량 = 63.55 g/mol 이므로,

0.00062916 mol × (63.55 g/mol) = 0.039983 g Cu^2+

 

(0.039983 g / 0.4225 g) × 100 = 9.463% Cu^2+

 

 

 

CuCO3의 몰질량 = 123.55 g/mol 이므로,

0.00062916 mol × (123.55 g/mol) = 0.077733 g CuCO3

 

(0.077733 g / 0.4225 g) × 100 = 18.40% CuCO3