a student is given 1.4799 g of pure cuo. to recover the cu present in the compound, the dark powdery solid was dissolved in 15.0 ml of 6 m hcl and the solution was diluted to 50.0 ml with water. how many grams of mg is needed to displace all of the copper (ii) ions from the solution?

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12-12-2022
Chemistry

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a student is given 1.4799 g of pure cuo. to recover the cu present in the compound, the dark powdery solid was dissolved in 15.0 ml of 6 m hcl and the solution was diluted to 50.0 ml with water. how many grams of mg is needed to displace all of the copper (ii) ions from the solution?

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SrikanthPMR SrikanthPMR · Answer 1 · 2022-12-17T17:00:26+01:00

It takes 0.4402 grams of magnesium to remove every copper (II) ion from the solution.

A substance's mole is equivalent to 6.022 x 1023 of that material (such as atoms, molecules, or ions). The term "Avogadro's number" or "Avogadro's constant" refers to the number 6.022 1023.

Calculation:

1 mole of CuO has 1 mole of ions, then 0.01834 moles of CuO will have 0.01834 moles of

Mg(s) + Cu² (aq) → Cu(s) + Mg²(aq)

According to reaction , 1 mole of copper (II) ion reacts with 1 mole of magnesium solid .

Then 0.01834 moles of copper (II) ion will react with :

1 × 0.01834 mol = 0.01834 mol of magnesium

Mass of 0.01834 moles of magnesium :

0.01834 mol × 24 g/mol = 0.4402 g

0.4402 grams of Mg is needed to displace all of the copper (II) ions from the solution.

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