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The following precipitation reaction can be used to determine the amount of copper ions dissolved in solution. A chemist added 5.00 × 10−2 L of a solution containing 0.173 mol L−1 Na3PO4(aq) to a 5.00 × 10−2 L sample containing CuCl2(aq). This resulted in a precipitate. The chemist filtered, dried, and weighed the precipitate. If 1.21 g of Cu3(PO4)2(s) were obtained, and assuming no copper ions remained in solution, calculate the following: the concentration of Cu2+(aq) ions in the sample solution. the concentrations of Na+(aq), Cl−(aq), and PO43−(aq) in the reaction solution (supernatant) after the precipitate was removed.

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From the stoichiometry of the reaction, 0.094 g  of [tex]Cu^{2+}[/tex] is contained in the reaction.

What is the amount of the copper ion?

The reaction equation is given by; [tex]2Na_{3}PO_{4} (aq) + 3CuCl_{2} (aq) ----- > 6NaCl (aq) + Cu_{3}(PO_{4}) _{2} (s)[/tex]. We must note that the balanced reaction equation is the entry way into obtaining the correct stoichiometry of the reaction in order to solve the problem.

Number of moles of [tex]Na_{3}PO_{4}[/tex] =  5.00 × 10−2 L *  0.173 mol L−1 = 0.00865 moles

If 2 moles of [tex]Na_{3}PO_{4}[/tex] produces 1 mole of[tex]Cu_{3}(PO_{4}) _{2}[/tex]

0.00865 moles of [tex]Na_{3}PO_{4}[/tex] produces  0.00865 moles * 1 mole/2 moles

= 0.0043 moles of [tex]Cu3(PO4)2[/tex]

Number of moles of [tex]Cu^{2+}[/tex] = 0.0043 moles/3 = 0.00144 moles

Mass of [tex]Cu^{2+}[/tex] = 0.00144 moles * 63.5 g/mol = 0.094 g

Learn more about stoichiometry:https://brainly.com/question/9743981

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