Question

A student is given 1.4583 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? Mg(s) + Cu2+(aq) --> Cu(s) + Mg2+(aq)

Answer 1

Answer:

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

Explanation:

Moles of CuO = $\frac{1.4583 g}{79.5 g/mol}=0.01834 mol$

1 mole of CuO has 1 mole of $Cu^{2+}$ ions, then 0.01834 molesof CuO will have 0.01834 moles of

$Mg(s) + Cu^{2+}(aq)\rightarrow Cu(s) + Mg^{2+}(aq)$

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 :

$\frac{1}{1}\times 0.01834 mol=0.01834 mol$ 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.