Question

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:

2Cu²⁺(aq) + 5I

Answer 1

This question is fairly simple if you have stoichiometry downreally well.

The question gives you enough information to find the mols ofI₃^- that were produced in the reaction.

(0.02996L Na₂S₂O₃)*(0.021mols/L Na₂S₂O₃) =6.2916*10^-4 mol

Looking at the second reaction you have there, you can findthe mols of I₃^- by dividing the mols ofNa₂S₂O₃ by two. This isthen the number of mols produced by the first reaction. Youcan then use the first reaction to find the number of mols ofCu²⁺ that was originally used. Essentially...

3.1458*10^-4 molsI₃^- * (2 mols Cu²⁺/1 molI₃^-) = 6.2916*10^-4 molCu²⁺ (63.5g Cu/1 mol Cu) = 0.03995 gCu²⁺

^{This number is the g that originally reacted, despite thefact that you added 0.4225g of sample. To find the masspercent, divide your calculated number by the original sample andmultiply by 100.}

^{(0.03995/0.4225)*100 = 9.45%}

^{The second part uses the mols of Cu2+.Thistime, instead of multiplying the mols of Cu by the mass of Cu,multiply by the mass of CuCO₃. Use this number,divide by 0.4225 and multiply by 100. This should give youthe 18.40%.}

Hope this helps!